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,  NEW   PLANE   AND   SOLID 

GEOMETRY. 


BY 


a 

WOOSTER    WOODRUFF    BEMAX 

H 

PROFESSOR   OF   MATHEMATICS   IX   THE   UNIVERSITY   OF   MICHIGAN 


DAYID   EUGENE    SMITH 

PROFESSOR   OF   MATHEMATICS   IX   TEACHERS   COLLEGE 
COLUMBIA    UNIVERSITY,  NEW   YORK 


BOSTON,  U.S.A. 

GTXX   &   COMPANY,   PUBLISHERS 

C&e  ^tfjenarum  press 

190.°, 


Copyright,  1895, 1899,  1900,  by 
Wooster  Woodruff  Beman  and  David  Eugene  Smith 


ALL  RIGHTS   RESERVED 


PREFACE. 


Ix  presenting  a  revision  of  their  "  Plane  and  Solid  Geom- 
etry "  (Boston,  1895),  the  authors  feel  that  an  explanation  of 
its  distinctive  features  may  be  of  service  to  the  teacher. 

It  is  sometimes  asserted  that  we  should  break  away  from 
the  formal  proofs  of  Euclid  and  Legendre  and  lead  the  student 
to  independent  discovery,  and  so  we  find  text-books  that  give 
no  proofs,  others  that  give  hints  of  the  demonstrations,  and 
still  others  that  draw  out  the  demonstration  by  a  series  of 
questions  which,  being  capable  of  answer  in  only  one  way, 
merely  conceal  the  Euclidean  proof.  But,  after  all,  the 
experience  of  the  world  has  been  that  the  best  results  are 
secured  by  setting  forth  a  minimum  of  formal  proofs  as 
models,  and  a  maximum  of  unsolved  or  unproved  propositions 
as  exercises.  This  plan  has  been  followed  by  the  authors, 
and  the  success  of  the  first  edition  has  abundantly  justified 
their  action. 

There  is  a  growing  belief  among  teachers  that  such  of  the 
notions  of  modern  geometry  as  materially  simplify  the  ancient 
should  find  place  in  our  elementary  text-books.  With  this 
belief  the  authors  are  entirely  in  sympathy.  Accordingly 
they  have  not  hesitated  to  introduce  the  ideas  of  one-to-one 
correspondence,  of  anti-parallels,  of  negative  magnitudes,  of 
general  figures,  of  prismatic  space,  of  similarity  of  point 
systems,  and  such  other  concepts  as  are  of  real  value  in  the 
early  study  of  the  science.     All  this  has  been  done  in  a  con- 

iii 


iv  rUEFACE. 

servative  way,  and  such  material  as  the  first  edition  showed 
to  be  at  all  questionable  has  been  omitted  from  the  present 
revision. 

Within  comparatively  recent  years  the  question  of  methods 
of  attack  has  interested  several  leading  writers.  Whatever 
has  been  found  to  be  usable  in  elementary  work  the  authors 
have  inserted  where  it  will  prove  of  most  value.  To  allow  the 
student  to  grope  in  the  dark  in  his  efforts  to  discover  a  proof, 
is  such  a  pedagogical  mistake  that  this  innovation  in  American 
text-books  has  been  generally  welcomed.  Upon  this  point 
the  authors  have  freely  drawn  from  the  works  of  Petersen  of 
Denmark,  and  of  Eouche  and  de  Comberousse  of  France,  and 
from  the  excellent  treatise  recently  published  by  Hadamard 
(Paris,  1898). 

With  this  introduction  of  modern  concepts  has  necessarily 
come  the  use  of  certain  terms  and  symbols  which  may  not 
generally  be  recognized  by  teachers.  These  have,  however, 
been  chosen  only  after  most  conservative  thought.  None  is 
new  in  the  mathematical  world,  and  all  are  recognized  by  the 
leading  writers  of  the  present  time.  They  certainly  deserve 
place  in  our  elementary  treatises  on  the  ground  of  exactness, 
of  simplicity,  and  of  their  general  usage  in  mathematical 
literature. 

The  historical  notes  of  the  first  edition  have  been  retained, 
it  being  the  general  consensus  of  opinion  that  they  add 
materially  to  the  interest  in  the  work.  For  teachers  who 
desire  a  brief  but  scholarly  treatment  of  the  subject  the 
authors  refer  to  their  translation  of  Fink's  "History  of 
Elementary  Mathematics"  (Chicago,  The  Open  Court  Pub- 
lishing Co.,  1899).  For  the  limitations  of  elementary  geom- 
etry, the  impossibility  of  trisecting  an  angle,  squaring  a 
circle,  etc.,   teachers  should   read  the  authors'  translation  of 


PREFA  CE.  V 

Klein's  valuable  work,  "Famous  Problems  of  Elementarv 
Geometry  "   (Boston.  Ginn  &  Company). 

It  is  impossible  to  make  complete  acknowledgment  of  the 
helps  that  have  been  used.  The  leading  European  text-books 
have  been  constantly  at  hand.  Special  reference,  however,  is 
due  to  such  standard  works  as  those  of  Henrici  and  Treutlein, 
"  Lehrbuch  der  Elementar-Geometrie,"  the  French  writers 
already  mentioned,  and  the  noteworthy  contributions  of  the 
recent  Italian  school  represented  by  Faifofer,  by  Socci  and 
Tolomei,  and  by  Lazzeri  and  Bassani. 

Teachers  are  urged  to  consider  the  following  suggestions  in 
using  the  book : 

1.  Make  haste  slowly  at  the  beginning  of  plane  and  of  solid 
geometry. 

2.  Never  attempt  to  give  all  of  the  exercises  to  any  class. 
Two  or  three  hundred,  selected  by  the  teacher,  should  suffice. 

3.  Eequire  frequent  written  work,  thus  training  the  eye,  the 
hand,  and  the  logical  faculty  together.  The  authors'  Geometry 
Tablet  (Ginn  &  Company)  is  recommended  for  this  work. 

W.  W.  BEMAN,  Anx  Arbor,  Mich. 
D.   E.   SMITH,  Brockport,  N.  Y. 
June  15,  1899. 


CONTENTS. 


INTRODUCTION. 

PAGE 

1.  Elementary  Definitions  .......         1 

2.  The  Demonstrations  of  Geometry  .....         9 

3.  Preliminary  Propositions         .......       13 


PLAXE    GEOMETRY. 

Book  I.  —  Rectilixear  Figures. 

1.  Triangles .21 

2.  Parallels  and  Parallelograms   ......   43 

3.  Problems  ...........   67 

4.  Loci  of  Points 80 


Book  II.  —  Equality  of  Polygons. 

1.  Theorems 90 

2.  Problems    .         .  ■ 109 

3.  Practical  Mensuration 112 


Book  III.  —  Circles. 

Definitions 114 

1.  Central  Angles        .........  116 

2.  Chords  and  Tangents 119 

3.  Angles  formed  by  Chords,  Secants,  and  Tangents     .         .  128 

4.  Inscribed    and    Circumscribed   Triangles    and    Quadrilat- 

erals    .  136 

5.  Two  Circles       ..........  143 

6.  Problems 146 

vii 


viii  CONTENTS. 

Appendix  to  Book  III. 

PAGE 

Methods 152 

Book  IV.  —  Ratio  and  Proportion. 

1.  Fundamental  Properties 150 

2.  The  Theory  of  Limits      ........  167 

3.  A  Pencil  of  Lines  Cut  by  Parallels 170 

4.  A  Pencil  Cut  by  Antiparallels  or  by  a  Circumference  .  177 

5.  Similar  Figures         .........  182 

6.  Problems    ...........  104 

Book  V.  —  Mensuration    of    Plane   Figures. 
Regular  Polygons  and  the  Circle. 

1.  The  Mensuration  of  Plane  Figures 100 

2.  The  Partition  of  the  Perigon 205 

3.  Regular  Polygons 200 

4.  The  Mensuration  of  the  Circle 216 

Appendix  to  Plane  Geometry. 

1.  Supplementary  Theorems  in  Mensuration    ....  226 

2.  Maxima  and  Minima 220 

3.  Concurrence  and  Collinearity 238 


SOLID   GEOMETRY. 

Book  VI.  —  Lines  and  Planes  in  Spacf 

1.  The  Position  of  a  Plane  in  Space.     The  Straight  Li 

the  Intersection  of  two  Planes 

2.  The  Relative  Position  of  a  Line  and  a  Plank 

3.  Pencil  of  Planes       ...... 

4.  Polyhedral  Angles  ..... 

5.  Problems    ........ 


NK    AS 


244 
251 
265 

274 
280 


COX  TENTS. 


Book    VII.  POLYHEDRA. 


1. 

General  and  Regular  Polyhedra           . 

283 

2. 

Parallelepipeds         ........ 

288 

3. 

Prismatic  and  Pyramidal  Space.     Prisms  and  Pyramids     . 

291 

4. 

The  Mensuration  of  the  Prism 

298 

5. 

The  Mensuration  of  the  Pyramid          . 

Book  VIII.  —  The  Cylinder,  Cone,  and  Sphere 
Similar  Solids. 

308 

1. 

The  Cylinder    .......... 

317 

2. 

The  Cone   

321 

■j. 

The  Sphere         

326 

4. 

The  Mensuration  of  the  Sphere    ...... 

355 

5. 

Similar  Solids  

364 

TABLES. 


Numerical  Tables 
Biographical  Table 
Table  of  Etymologies  . 
Index        . 


371 
372 
375 
379 


PLANE  AND  SOLID  GEOMETKY. 


3>^C 


PLANE  GEOMETRY. 

INTRODUCTION. 
1,    ELEMENTARY   DEFINITIONS. 

1.  In  Arithmetic  the  student  has  considered  the  science  of 
numbers,  and  has  found,  for  example,  that  a  number  which 
ends  in  5  or  0  is  divisible  by  5. 

In  Algebra  he  has  studied,  among  other  things,  the  equation, 
and  has  found  that  if  J  x  —  1  —  5,  x  must  equal  12. 

In  Geometry  he  is  to  study  form,  and  he  will  find,  for 
example,  that  two  triangles  must  necessarily  be  equal  if  the 
three  sides  of  the  one  are  respectively  equal  to  the  three  sides 
of  the  other. 

Before  beginning  the  subject,  however,  there  are  certain 
terms  which,  although  familiar,  are  used  with  such  exactness 
as  to  require  careful  explanation.  These  terms  are  solid,  sur- 
face, line,  angle  (with  various  kinds  of  each),  and  point.  As 
with  most  elementary  mathematical  terms,  such  as  number, 
space,  etc.,  it  is  difficult  to  give  them  simple  and  satisfactory 
definition.  Explanations  can,  however,  be  given  which  will 
lead  the  student  to  a  reasonable  understanding  of  them. 

2.  The  space  with  which  we  are  familiar  and  in  which  we 
live  is  evidently  divisible.  Any  limited  portion  of  space  is 
called  a  solid. 

In  geometry  no  attention  is  given  to  the  substance  of  which 
the  solid  is  composed.     It  may  be  water,  or  iron,  or  air,  or 


2  PLANE   GEOMETRY.  [Intr. 

wood,  or  it  m?y  be  a  vacuum.  Indeed,  geometry  considers 
only  the  space  occupied  by  the  substance.  This  space  is  called 
a  geometric  solid,  or  simply  a  solid,  while  the  substance  is  called 
a  physical  solid.  Thus,  a  ball  is  a  physical  solid ;  the  space 
which  the  ball  occupies  is  a  geometric  solid. 

3.  That  which  separates  one  part  of  space  from  an  adjoin- 
ing part  is  called  a  surface.  So  we  speak  of  the  surface  of  a 
ball,  the  surface  of  the  earth,  etc. 

4.  Every  surface  is  divisible.  That  which  separates  one 
part  of  a  surface  from  an  adjoining  part  is  called  a  line. 

5.  Every  line  is  divisible.  That  which  separates  one  part 
of  a  line  from  an  adjoining  part  is  called  a  point. 

A  point  is  not  divisible. 

Thus,  in  the  figure  the  surface  of  the  block  separates  the  space  occu- 
pied by  the  block  from  all  the  rest  of  space.     This  surface  is  divisible  in 
many  ways ;   for  example,  it  is  divided  into 
two  parts  by  the  line  passing  from  A  through 

§ B  and  C  and  back  to  A.     This  line  is  divisible 

\      in  many  ways ;   for  example,  it  is  separated 

\    into  three  parts  by  the  points  A,  B,  C.    In  the 

case  of  a  line  that  returns  into  itself, — i.e.  a 
closed  line,  like  the  one  just  mentioned,  —  two  points  are  necessary  com- 
pletely to  separate  one  part  from  the  other. 

It  is  impossible  to  draw  mechanically  a  geometric  line.  A 
chalk  mark,  a  thread,  a  fine  wire,  an  ink  mark,  are  all  very 
thin  physical  solids  used  to  represent  lines ;  for  this  purpose 
they  are  very  helpful.  So,  too,  a  dot  may  be  used  to  represent 
a  point,  and  a  sheet  of  paper  may  be  used  to  represent  a  surface, 
although  each  is  really  a  physical  solid. 

6.  The  preceding  definitions  start  from  the  solid  and  take 
the  surface,  line,  and  point  in  order.  It  is  also  possible  to 
start  with  the  point  and  proceed  in  reverse  order. 

The  point  is  the  simplest  geometric  concept ;  it  lias  position, 
but  not  magnitude. 


Sec.  7.]  ELEMENTARY  DEFINITIONS.  3 

A  moving  point  describes  a  line. 

This  may  be  represented  by  a  pencil  point  moving  on  a  piece  of  paper. 

A  moving  line  describes,  in  general,  a  surface. 

This  may  be  represented  by  a  crayon  lying  flat  against  the  blackboard, 
and  moving  sidewise.  How  may  a  line  move  so  as  not  to  describe  a 
surface  ? 

A  moving  surface  describes,  in  general,  a  solid. 

Thus,  the  surface  of  a  glass  of  water,  as  it  moves  upward,  may  be  said  to 
describe  a  solid.    How  may  a  surface  move  so  as  not  to  describe  a  solid  ? 

7.  Through  two  points  any  number  of  lines  may  be  imagined 
to  pass. 

For  example,  through  the  points  Pi,  I 
(read  "  P-one,  P-two")  the   lines  q,  r, 
may  be  imagined  to  pass. 

A  straight  line  is  a  line  which  is  determined  by  any  two  of 
its  points. 

In  the  figure,  s  represents  a  straight  line,  for,  given  the  points  Pu  P2 
on  the  line,  its  position  is  fixed  ;  it  is  determined. 

But  q  and  r  do  not  represent  straight  lines,  because  Pi  and  P2  do  not 
determine  them. 

The  word  line,  used  alone,  is  to  be  understood  to  refer  to  a 
straight  line. 

The  expression  straight  line  is  used  to  mean  both  an  unlim- 
ited straight  line  and  a  portion  of  such  a  line.  In  case  of 
doubt,  line-segment,  or  merely  segment,  is  used  to  mean  a 
limited  straight  line. 

As  has  been  seen,  a  point  is  usually  named  by  some  capital 

letter.      A    segment    is    usually 

A        B                C 
named  by  naming  its  end  points,  ! ■ 1 ■    0   ■ — 

or  by  a  single  small  letter. 

In  the  annexed  figure,  AB,  AC,  BC,  and  o  are  marked  off. 

Two  segments  are  said  to  be  equal  when  they  can  be  made 
to  coincide. 


4  PLANE    GEOMETRY.  [Intr. 

8.  If  three  points,  A,  B,  C,  are  taken  in  order  on  a  line, 
as  in  the  preceding  figure,  then  the  line-segment  AC  is  called 
the  sum  of  the  line-segments  AB  and  BC,  and  AB  is  called  the 
difference  between  AC  and  BC. 

9.  If  a  point  divides  a  line-segment  into  two  equal  seg- 
ments, it  is  said  to  bisect  the  line-segment  and  p 

is  called  its  mid-point. 

A  line  is  easily  bisected  by  the  use  of  a  straight- 
edge and  compasses,  thus  : 

With  centers  A  and  B,  and  equal  radii,  describe 
arcs  intersecting  at  P  and  P'.  A~ 

Draw  PP'.     This  bisects  AB. 

The  proof  of  this  fact  is  given  later. 

10.  If  a  segment  is  drawn  out  to  greater 
length,  it  is  said  to  be  produced.  'p^ 

To  produce  AB  means  to  extend  it  through  B,  toward  C,  in  the  second 
figure  in  §  7.    To  produce  BA  means  to  extend  it  through  A,  away  from  B. 

11.  A  line  not  straight,  but  made 
up  of  straight  lines,  is  called  a  broken 
line. 

12.    Through  three  points,  not  in  a  straight  line,  any  num- 
ber   of   surfaces   may  be  imagined 
to  pass. 

For  example,  through  the  points  A,  B, 
C  the  surfaces  P  and  S  may  be  imagined 
to  pass. 

A  plane  surface  (also  called  a  plane)  is  a  surface  which  is 
determined  by  any  three  of  its  points  not  in  a  straight  line. 

In  the  figure,  P  represents  a  plane,  for  it  is  determined  by  the  points 
A,  B,  C.     But  S  does  not  represent  such  a  surface. 

A  plane  is  indefinite  in  extent  unless  the  contrary  is  stated. 
To  produce  it  means  to  extend  it  in  length  or  breadth. 


Secs.  13,  14.]  ELEMENTARY  DEFINITIONS.  5 

13.  If  two  lines  proceed  from  a  point,  they  are  said  to  form 
an  angle,  the  lines  being  called  the  arms,  and  the  point  the 
vertex,  of  that  angle. 

The  size  of  the  angle  is  independent  of  the  length  of  the 
arms ;  the  size  depends  merely  upon  the  amount  of  turning 
necessary  to  pass  from  one  arm  to  the  other. 

The  methods  of  naming  an  angle  will  be  seen  from  the 
annexed  figures.  It  is  convenient  to  letter  an  angle  around 
the  vertex,  as  indicated  by  the  arrows,  that  is,  opposite  to  the 
course  of  clock-hands,  or  counter-clockwise. 


C 


-A 


\  ^ 

Angle  m.  Angle  0.  Angle  A  OB.        Angle  a  b.  Angle  AOB. 

A  line  proceeding  from  the  vertex,  turning  about  it  counter- 
clockwise from  the  first  arm  to  the  second,  is  said  to  turn 
through  the  angle,  the  angle  being  greater  as  the  amount  of 
turning  is  greater. 

14.  If  the  two  arms  of  an  angle  lie  in  the  same  straight 
line  on  opposite  sides  of  the  vertex,  a  straight  angle  is 
said  to  be  formed.  If  the  angle  still  further  increases,  until 
the  moving  arm  has  performed  a  complete  revolution,  thus 
passing  through  two  straight  angles,  a  perigon  is  said  to  be 
formed. 

For  practical  purposes  angles  are  measured  in  degrees,  min- 
utes,   and    seconds.      A 

perigon  is   said  to  con-    b  £  0 

tain  360°.  °  A 

In  general,  if  tWO  lines     A0B'  a  stT&iS^t  angle.  A  perigon,  or  angle 

_  BOA,  a  straight  angle.  of  360°. 

are  drawn  irom  O,  two 

angles,  each  less  than  a  perigon,  are  formed.  Of  these  the 
smaller  is  always  to  be  understood  if  "the  angle  at  0"  is 
mentioned,  unless  the  contrary  is  stated. 


6 


PLANE   GEOMETRY. 


[Lntr. 


15.  If  a  line  turns  through  an  angle,  all  points  or  line- 
segments  through  which  it  passes  in  its  turning,  except  the 
vertex,  are  said  to  be  within  the  angle.  Other  points  or  lines 
are  either  on  the  arms  or  without  the  angle. 

16.  Two  angles,  ab,  a'b',  are  said  to  be  equal  when,  without 
changing  the  relative  position  of  a  and  b,  angle  ab  may  be 
placed  so  that  a  lies  along  a',  and  b  along  b'. 

This  equality  is  tested  by  placing  one  angle  on  the  other,  the  vertices 
coinciding.  Then  if  the  arms  can  he  made  to  coincide,  the  angles  are 
equal,  otherwise  not. 

17.  If  three  lines,  OA,  OB,  OC,  proceed  from  a  common 
point  0,  OB  lying  within  the  angle  AOC, 
then  angles  A  OB  and  BO C  are  called  ad- 
jacent angles.  Angle  AOC  is  called  the  sum 
of  the  angles  A  OB,  BOC.  Either  of  the 
adjacent  angles  is  called  the  difference  be- 
tween angle  AOC  and  the  other  of  the  adjacent  angles. 

As  two  angles  may  be  added,  so  several  may  be  added. 

18.  If  a  line  divides  an  angle  into  two  equal  angles,  it  is 
said  to  bisect  the  angle  and  is  called  its 
bisector. 

In  the  annexed  figure,  if  angle  A  OY  equals 
angle  YOB,  then  OY  is  the  bisector  of  angle 
A  OB. 

And,  in  general,  to  bisect  any  magnitude  means 
to  divide  it  into  two  equal  parts. 

An  angle  is  easily  bisected  by  the  use  of  a 
straight-edge  and  compasses,  thus : 

If  AOB  is  the  given  angle,  mark  off  with  the 
compasses  OC  equal  to  OD. 

Then  with  C  and  D  as  centers  and  CD  as  a 
radius  draw  two  arcs  intersecting  at  P  and  P'. 

The  line  joining  PorF  with  0  is  the  required  bisector.  The  proof 
of  this  fact  is  given  later. 


Secs.  19-22.]  ELEMENTARY  DEFINITIONS  7 

19.  A  right  angle  is  half  of  a  straight  angle, 

It  follows  from  this  definition  that  the  sum  of  two  rigid 
angles  is  a  straight  angle;  and  from  the 
definitions   of   a   straight   angle   and  of  a 
perigon,  that  the  sum  of  two  straight  an- 
gles, or  of  four  right  angles,  is  a  perigon.      C 

It    also    follows    that  a   straight    angle 
contains  180°  and  a  right  angle  contains  90°. 

20.  If  two  lines  meet  and  form  a  right  angle,  each  line  is 
said  to  be  perpendicular  to  the  other. 

Each  is  also  spoken  of  as  a  perpendicular  to  the  other. 
Thus,  in  the  preceding  figure,  BO  is  perpendicular  to  CA,  or 
is  a  perpendicular  to  CA.  The  segment  PO  is  called  the  per- 
pendicular from  P  to  CA,  since  it  will  presently  be  proved  that 
it  is  unique  ;  that  is,  that  there  is  one  and  only  one  perpendic- 
ular.     0  is  called  the  foot  of  that  perpendicular. 

The  word  unique,  meaning  one  and  only  one,  is  frequently 
used  in  mathematics. 

A  line  is  easily  drawn  perpendicular  to  another  line  by  the  use  of  a 
straight-edge  and  compasses.  This  is  seen  in  the  figure  in  §  9,  where 
PP"  is  perpendicular  to  A  B. 

21.  An  angle  less  than  a  right  angle  is  said  to  be  acute ;  one 
greater  than  a  right  angle  but  less  than  a  straight  angle  is  said 
to  be  obtuse ;  one  greater  than  a  straight  angle  but  less  than  a 
perigon  is  said  to  be  reflex  or  convex. 

22.  Two  lines  which  form  an  acute,  obtuse,  or  reflex  angle 
are  said  to  be  oblique  to  each  other. 

Acute,  obtuse,  and  reflex  angles  are  classed  under  the  gen- 
eral term  oblique  angles. 

The  meaning  of  the  expressions  oblique  lines,  an  oblique,  foot  of  an 
oblique,  will  be  understood  from  §  20. 

Draw  a  figure  representing  acute,  obtuse,  and  reflex  angles,  oblique 
lines,  an  oblique  from  P  to  CA,  the  foot  of  an  oblique. 


8  PLANE   GEOMETRY.  [Intr. 

23.  Two  angles  are  said  to  be  complements  of  each  other  if 
their  sum  is  a  right  angle.  Two  angles  are  said  to  be  supple- 
ments of  each  other  if  their  sum  is  a  straight  angle.  Two 
angles  are  said  to  be  conjugates  of  each  other  if  their  sum  is  a 
perigon. 

If  one  angle  is  the  complement  of  another,  the  two  angles 
are  said  to  be  complemental  or  comple- 
mentary.    Similarly,  if  one  angle  is  the 
supplement  of  another,  the  two  angles  are 
c_ / a  said  to  be  supplemental  or  supplementary. 

In  the  annexed  figure,  angles  AOB  and  BOC 
are  supplemental,  also  angles  BOC  and  COD, 
D  etc. 

24.  If  two  lines,  CA,  DB,  intersect  at  0,  as  in  the  above 
figure,  the  angles  A  OB  and  COD  are  called  vertical  or  opposite 
angles;  also  the  angles  BOC  and  DO  A. 


Exercises.  1.  How  many  degrees  in  a  right  angle  ?  How  many 
minutes  ?     How  many  seconds  ? 

2.  What  is  the  complement  of  one-half  of  a  right  angle  ?  of  one- 
fourth  ? 

3.  How  many  degrees  in  the  supplement  of  an  angle  of  (a)  75°  ? 
(b)  90°?    (c)  150°?    (d)  179°? 

4.  Also  in  the  complement  of  an  angle  of  (a)  75°  ?  (b)  1°  ?  (c)  89°  ? 
(d)45°?  (e)  90°?  (f)0°? 

5.  Also  in  the  conjugate  of  an  angle  of  (a)  270°  ?  (b)  180°  ?  (c)  359°  ? 
(d)  90°?  (e)  1°?  (f)  360°? 

6.  Draw  a  figure  showing  that  two  straight  lines  determine  one  point ; 
also  one  showing  that  three  straight  lines  determine,  in  general,  three 
points. 

7.  How  many  degrees  in  each  of  the  two  conjugate  angles  which  the 
hour  and  minute  hands  of  a  clock  form  at  4  o'clock  ? 

8.  If  six  lines,  proceeding  from  a  point,  divide  a  perigon  into  six 
equal  angles,  express  one  of  those  angles  (a)  in  degrees,  (b)  as  a  fraction 
of  a  right  angle,  (c)  as  a  fraction  of  a  straight  angle. 


Secs.  25-27.1        DEMONSTRATIONS    OF   GEOMETRY. 


2.     THE   DEMONSTRATIONS    OF   GEOMETRY. 

25.  The  object  of  geometry  is  the  investigation  of  truths  con- 
cerning combinations  of  lines  and  points,  and  of  the  methods 
of  making  certain  constructions  from  lines  and  points. 

26.  A  proposition  is  a  statement  of  either  a  truth  to  be 
demonstrated  or  a  construction  to  be  made. 

For  example,  geometry  investigates  this  proposition  :  If  two  lines 
intersect,  the  vertical  angles  are  equal.  It  also  investigates  the  methods 
of  drawing  a  line  perpendicular  to  another  line,  and  various  other  propo- 
sitions requiring  some  construction. 

Propositions  are  divided  into  two  classes  —  theorems  and 
problems. 

A  theorem  is  a  statement  of  a  geometric  truth  to  be  demon- 
strated. 

A  problem  is  a  statement  of  a  geometric  construction  to  be 
made. 

For  example  :  Theorem,  If  two  lines  intersect,  the  vertical  angles  are 
equal.  — Problem,  Required  through  a  point  in  a  line  to  draw  a  perpen- 
dicular to  that  line. 

27.  There  are  a  few  geometric  statements  so  obvious  that  the 
truth  of  them  may  be  taken  for  granted,  and  a  few  geometric 
operations  so  simple  that  it  may  be  assumed  that  they  can  be 
performed.  Such  a  statement,  or  the  claim  to  perform  such 
an  operation,  is  called  a  postulate. 

The  geometric  operations  thus  assumed  require  the  use  of  the 
straight-edge  and  compasses.  The  straight-edge  and  the  compasses  are 
the  only  instruments  recognized  in  elementary  geometry. 

The  postulates  used  in  this  work  are  set  forth  from  time  to 
time  as  required.  At  present  three  general  classes  suffice, 
as  follows : 


10  PLANE    GEOMETRY.  [Intr. 

28.  Postulates  of  the  Straight  Line. 

1.  Two  points  determine  a  straight  line. 
This  follows  from  the  definition. 

2.  Two  straight  lines  in  a  plane  determine  a  point. 

3.  A  straight  line  may  be  drawn  and  revolved  about  one 
of  its  points  as  a  center  so  as  to  include  any  assigned  point 
in  space. 

4.  A  straight  line-segment  may  be  produced. 

5.  A  straight  line  is  divided  into  two  parts  by  any  one  of 
its  points. 

29.  Postulates  of  the  Plane. 

1.  Three  points  not  in  a  straight  line  determine  a  plane. 
This  follows  from  the  definition. 

2.  A  straight  line  through  two  points  in  a  plane  lies  wholly 
in  the  plane. 

Thus,  if  part  of  a  straight  line  lies  in  an  unlimited  plane  blackboard, 
the  whole  line  lies  in  the  blackboard. 

3.  A  plane  may  be  passed  through  a  straight  line  and  re- 
volved about  it  so  as  to  include  any  assigned  point  in  space. 

4.  A  portion  of  a  plane  may  be  produced. 

5.  A  plane  is  divided  into  two  parts  by  any  one  of  its 
straight  lines,  and  space  is  divided  into  two  funis  by  any 
plane. 

30.  Postulate  of  Angles. 

All  straight  angles  are  equal. 

31.  There   arc  also  a  number  of    simple   statements,  of  a 

general    nature,   so   obvious    that-    the   truth    of    them    may   be 

taken   for  granted.     These  are  called  axioms. 

The  following  arc  the  axioms  mosl  frequently  used  in  geometry,  and 
they  arc  so  important  that  they  should  be  learned  by  number. 


Sec.  32.]  DEMONSTRATIONS    OF   GEOMETRY.  11 

32.    Axioms. 

1.  Tilings  which   are  equal  to  the  same  thing,  or  to  equal 

things,  are  equal  to  each  other. 

That  is,  (1)  if  A  =  JJ,  and  C  =  B,  then  4  =  C.     Or,  (2)  if  J.  =  B,  and 

7>  =  C,  and  C  =  D,  then  J.  =  Z). 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 
That  is,  if  A  =  B,  and  if  C  =  B.  then  A  +  C  =  B  +  D. 

f>.  If  equals  are  subtracted  from  equals,  the  remainders  are 
equal. 

That  is,  if  A  =  B,  and  if  C  =  1),  then  A  -  C  =  B  -  D. 

^4.    If  equals  ore  added  to  unequals,  the  sums  are  unequal  in 
the  same  sense. 

That  is,  if  A  =  B.  and  if  C  is  greater  than  D,  then  A  +  C  is  greater 
than  B  ~  I). 

■h^ If  equals  are  subtracted  from  unequals,  the  remainders 
are  unequal  in  the  same  sense. 

That  is,  if  ^4  =  B,  and  if  C  is  greater  than  D,  then  C  —  A  is  greater 
than  D  -  B. 

6.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

That  is,  if  A  =  B.  and  m  is  any  number,  then  mA  =  mB. 

s 

7.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 

A      B 

That  is,  as  in  axiom  6,  —  =—  •     It  will  be  seen  that  axiom  6  covers 
m      m 

axiom  7.  for  m  may  be  a  fraction. 

8.  The  whole  is  greater  than  any  of  its  parts,  and  equals 
the  sum  of  all  its  parts. 

The  latter  part  of  this  axiom  is  merely  the  definition  of  whole. 

9.  If  three  magnitudes  are  so  related  that  the  first  is  greater 
than  the  second,  while  the  second  is  greater  than,  or  equal  to, 
the  third,  then  the  first  is  g  renter  than  the  third. 

E.g.  if  A  is  greater  than  7?.  and  if  B  is  greater  than,  or  equal  to,  C, 
then  A  is  greater  than  C. 


12 


PLANE    GEOMETRY. 


[Intr. 


33. 


Symbols  and  Abbreviations. 


The  following  are  used  in  this  work,  and  are  inserted  here 
merely  for  reference,  and  not  for  memorizing  : 


e.g.  Latin,  exempli  gratia,  for 

example. 
i.e.  Latin,  id  est,  that  is. 

since. 

therefore, 
pt.,  pts.    point,  points, 
rt.  right, 

st.  straight. 

ax.  axiom, 

post.         postulate, 
def.  definition, 

prop.         proposition, 
th.  theorem, 

pr.  problem, 

cor.  corollary, 

subst.        substitution, 
prel.  preliminary, 

const.        construction, 
ppd.  parallelepiped. 

^  arc. 

O,  (s)       circle,  circles. 
A,  A      triangle,  triangles. 
□  ,  Q[]       square,  squares. 
I    I.  \j~\     rectangle,  rectangles. 
O,  UJ    parallelogram,   parallelo- 
grams. 
Z.,  A        angle,  angles. 
+  pins,  increased  by. 

—  minus,  diminished  by. 

X ,  •  ,       and  absence  of  sign,  de- 
note multiplication. 


/. 


de- 


> 
< 

> 

< 


and  fractional  form, 
note  division. 

is  equal,  or  equivalent,  to. 

is  identical  with,  as  AB  =  AB, 
or  coincides  with. 

is  congruent  to. 

is  similar  to. 

approaches  as  a  limit. 

is  greater  than. 

is  less  than. 

is  not  equal  to,  i.e.  >  or  < . 

is  not  greater  than,  i.e.  =  or  <  . 

is  not  less  than,  i.e.  =  or  >. 

is  perpendicular  to,  or  a  per- 
pendicular. 

is  parallel  to,  or  a  parallel. 

and  so  on. 


The  above  take  the  plural  also; 
thus,  =  means  are  equal,  as  well 
as  is  equal. 

The  manner  of  reading  some  of  the 
familiar  symbols  is  suggested,  as 
follows : 

P',  P-prime;   P",  P-second ;  P'", 

P-third,  etc. 
P1?  P-one  ;  P2,  P-two,  etc. 
A'B',  A-prime  B-prime,  etc. 
A{,  A-one-prime,  etc. 


References  to  preceding  propositions  are  made  by  book  and  proposi- 
tion thus,  I,  prop.  IV  ;  if  the  first  Roman  numeral  is  omitted,  the  prop- 
osition is  in  the  current  book.     Section  references  are  also  used. 

Other  simple  abbreviations  are  occasionally  used,  but  they  will  be 
easily  understood. 


Prop.  I.] 


PRELIMINARY  PROPOSITIONS. 


13 


3.     PRELIMINARY  PROPOSITIONS. 

34.  The  following  theorems  are  designed  to  show  to  the 
beginner  the  nature  of  a  geometric  proof,  and  to  lead  him  by 
easy  steps  to  appreciate  the  logic  of  geometry.  Some  of  them 
might  properly  have  been  incorporated  in  Book  I,  and  others 
might  have  been  omitted  altogether ;  but  they  form  a  group  of 
simple  propositions  which  lead  the  student  up  to  the  more  diffi- 
cult work  of  geometry,  and  for  that  reason  they  are  inserted 
here.  The  student  and  the  teacher  are  advised  to  proceed 
slowly  until  the  logic  of  the  subject  is  understood,  and  under 
no  circumstances  to  allow  mere  memorizing  of  the  proofs. 


Proposition  I. 

35.    Theorem,      All  right  angles  are  equal. 

Suggestion,  The  only  angles  of  whose  equality  we  are  thus  far 
assured  are  straight  angles.  Hence  in  some  way  we  must  base  our  proof 
of  this  theorem  on  the  postulate  of  angles,  which  asserts  this  fact.  We 
then  consider  how  a  right  angle  is  related  to  a  straight  angle,  and  the 
proof  is  at  once  suggested. 


any  two  right  angles,  r,  r\ 
that  r  =  r'. 


Given 

To  prove 

Proof,    1.      r  and  r  are  halves  of  straight  angles.  Def .  rt.  Z 

(§  19.    A  right  angle  is  half  of  a  straight  angle.) 

2.  All  straight  angles  are  equal.  §  30 

3.  .'.  all  right  angles,  and  hence  r  and  ?•',  are  equal. 

Ax.  7 

(If  equals  are  divided  by  equals,  the  quotients  are  equal.) 


14 


PLANE    GEOMETRY. 


[Intr. 


Proposition  II. 

36.  Theorem.  At  a  given  point  in  a  given  line  not  more 
than  one  perpendicular  can  be  drawn  to  that  line  in  the  same 
plane. 

Y    ,.7 


V 


Given  YYf  _L  XX  at  0. 

To  prove     that  no  other  perpendicular  can  be  drawn  to  XX\ 
at  0,  in  the  same  plane. 

Proof.    1.    Suppose  that  another  _L,  ZZ',  could  be  drawn. 

2.  Then  ZXOZ  would  be  a  rt.  Z.  Def .  ± 

(If  two  lines  meet  and  form  a  rt.  Z,  each  is  said  to  be  ±  to  the  other. ) 

3.  But  Z  XOY  is  a  rt.  Z.  Given  ;  def.  _L  §  20 
(For  it  is  given  that  YY'  _L  XX',  and  the  def.  of  a  _L  is  given  in  step  2.) 

4.  .-.  Z  XO  Y  would  equal  Z  XOZ.  Prop.  I 

(All  right  angles  are  equal.) 

5.  J>ut  this  is  impossible.  Ax.  8 
(The  whole  is  greater  than  any  of  its  parts,  etc.) 

6.  .'.  the  supposition  of  step  1  is  absurd,  and  a  second 
perpendicular  is  impossible. 

Note.  In  prop.  I  we  proved  directly  from  the  definition  of  straight 
angle  that  all  right  angles  are  equal.  In  prop.  II  a  different  method  of 
proof  is  followed.  We  have  here  supposed  that  the  theorem  is  false  and 
have  shown  that  this  supposition  is  absurd.  Such  proofs  have  long  been 
known  by  the  name  "  reductio  ad  ahs/tnlnm,"  a  reduction  to  an  absurd- 
ity.    They  are  also  called  indirect  proofs. 


Props.  Ill,  IV.]         PRELIMINARY   PROPOSITIONS.  15 


Proposition  III. 

37.    Theorem.      The  complements  of  equal  angles  are  equal. 

Suggestion*.  Three  lines  of  proof  may  present  themselves.  We  may 
base  our  proof  on  the  equality  of  straight  angles,  as  we  did  in  prop.  I,  or 
we  may  take  an  indirect  proof  as  in  prop.  II,  beginning  by  supposing  the 
theorem  false  and  showing  the  absurdity  of  this  supposition,  or  we  may 
base  the  proof  on  prop.  I.  Since  the  complements  suggest  right  angles, 
which  of  the  three  methods  would  it  probably  be  best  to  follow  ? 


0  A        0'  A' 

Given         two  equal  A,  AOB,  A'O'B',  and  their  complements. 
BOC,  B'O'C,  respectively. 

To  prove    that  ABOC  =  A  B'O'C. 

Proof.    1.  A  AOC  and  A'O'C  are  rt.  A.  Def.  compl. 

(§  23.    Two  A  are  said  to  be  complements  if  their  sum  is  a  rt.  Z.) 

2.  .'.  Z  AOC  =  Z  A'O'C.  Prop.  I 

(All  right  angles  are  equal.) 

3.  But  Z  AOB  =  Z  A'O'B'.  Given 

4.  .\  A  BOC  =  A  B'O'C  Ax.  3 
(If  equals  are  subtracted  from  equals,  the  remainders  are  equal.) 


Proposition  IV. 

38.    Theorem.      The  supplements  of  equal  angles  are  equal. 

Let  the  student  draw  the  figure  and  give  the  proof  after  the  manner  of 
prop.  III.     Use  only  four  steps  in  the  proof. 

Given 

To  prove 

Proof. 


16  PLANE   GEOMETRY.  [Intr. 

Proposition  V. 
39.    Theorem.      The  conjugates  of  equal  angles  are  equal. 


Given  two  equal  angles,  ab,  a'b'. 

To  prove    that  Aba  —  A  b'a'. 

Proof.    1.  The  given  A  may  be  so  placed  that  a  lies  along  a', 
and  b  along  b'.  Def.  equal  A 

(§  16.    Two  A,  ab,  a'b',  are  said  to  be  equal  when  Z  ab  can  be  placed 
so  that  a  lies  along  a',  and  6  along  b'.) 

2.  But  then  A  ba  must  equal  A  b'a'.  Def.  equal  A 


Proposition  VI. 

40.  Theorem.  If  two  lines  cut  each  other,  the  vertical 
angles  are  equal. 

Suggestion.  After  examining  the  figure  the  student  might  say  that 
because  Z  a  +  Z  b  =  st.  Z,  and  Z  b  +  Z  a'  =  st.  Z,  .-.  Z  a  +  Z  6  =  Z  6 
+  Z  a',  and  then  subtract  Z  6  from  these  equals  ;  or  he  might  say  that 
Za  —  Za'  because  each  is  the  supplement  of  Z  b.  He  should  always  feel 
encouraged  to  try  various  proofs,  selecting  the  shortest  and  the  clearest. 
Does  the  following  proof  meet  these  requirements  ? 

Given  two   lines    cutting    each    oilier, 

forming  two  pairs  of    opposite 
angles,  a,  a',  and  b,  b'. 

To  prove     that  A  a  —  A  a '. 

Proof.    1.  Aa  and  A  a'  are  supplements  of  Ab.        Def.  suppl. 
(§  23.    Two  A  are  said  to  be  supplements  if  their  sum  is  a  st.  Z.) 
2.  .\Aa  =  A«'.  Prop.  IV 

(The  supplements  <>1"  equal  angles  are  equal.) 


Props.  VII,  VIII.]       PRELIMINARY  PROPOSITIONS.  17 

Proposition  VII. 

41.    Theorem.     A  line-segment  can  be  bisected  in  only  one 
point. 

P 
A  ST  B 

Given  a  line-segment  AB,  bisected  at  M. 

To  prove     that  there  is  no  other  point  of  bisection. 

Proof.    1.   Suppose  another  point  of  bisection  exists,  as  P.  be- 
tween M  and  B. 

2.  Then  since  AM  and  AP  are  both  halves  of  A  P.  they 

are  equal.  Ax.  7 

(State  ax.  7.) 

3.  But  this  is  impossible,  for  AM  is  part  of  AP.     Ax.  8 

(State  ax.  8.) 

4.  .'.  the  supposition  that  there  is  a  second  point   of 
bisection  is  absurd. 

(Another  reductio  ad  absurdum,  as  in  prop.  II.) 


Proposition  VIII. 

42.    Theorem.     An  angle  can  be  bisected  by  only  one  line. 
(The  student  may  prove  this  after  the  manner  of  prop.  VII.) 


Exercises.     9.    Of  two  supplemental   angles,    a  and  b,    (a)   suppose 
a  —  2b,  how  many  degrees  in  each  ?    (b)  suppose  a  =  3 6,  how  many  ? 

10.  How  many  straight  lines   are,  in  general,  determined   by  three 
points?   by  four?     (The  points  in  the  same  plane.) 

11.  If  of  five  angles,  a,  b.  c,  d,  e.  whose  sum  is  a  perigon,  a  =  20°, 
b  =  30°,  c  —  40°,  d  =  50°,  how  many  degrees  in  e  ? 

12.  Of  three  angles  whose  sum  is  a  perigon,  the  first  is  twice  the  sec- 
ond, and  the  second  three  times  the  third;  how  many  degrees  in  each  ? 


18 


PLANE    GEOMETRY. 


[Intr. 


Proposition  IX. 

43.  Theorem.  The  bisectors  of  two  adjacent  angles  formed 
by  one  line  cutting  another  are  perpendicular  to  each  other. 

Suggestion.  Considering  the  figure,  we  see  that  to  prove  OA  _L  OB 
we  must  show  that  Z  A  OB  is  a  rt.  Z.  Now  the  only  way  that  we  have  as 
yet  of  showing  an  angle  to  be  a  right  angle  is  to  show  that  it  is  half  of  a 
straight  angle.  But  evidently  Z  A  OY  is  half  of  Z  XOY,  because  Z  XOY 
is  bisected ;  similarly,  Z  YOB  is  half  of  Z  YOX',  and  this  suggests  the 
following  proof. 


Given  two  lines,  XX',  YY',  cutting  at   0 ;  also  OA,  OB, 

bisecting  A  XOY,  YOX'.  respectively. 


that  OA  _L  OB. 
AAOY  =  \AXOY. 
Z  YOB  =  ±A  YOX'. 


To  prove 

Proof.    1. 
2. 

3.  ;.ZAOB  =  \AXOX'. 

(If  equals  are  added  to  equals,  the  sums  are  equal.) 

4.  .-.  Z  AOB  =  i  of  a  st.  Z.  Def.  st.  Z 

(§  14.    If  the  two  arms  of  an  Z  lie  in  the  same  st.  line  on  opposite  sides 
of  the  vertex,  a  st.  Z  is  said  to  be  formed.) 


Given;  §  18 

(liven;   §  18 
Ax.  2 


5.  .-.  Z  AOB  =  ait.  Z. 

(§  19.    A  rt.  Z  is  half  of  a  st.  Z.) 

6.  .-.  0A±  OB. 


Def.  rt.  Z 


Def.  _L 


(§  20.    If  two  lines  meet  and  form  a  rt.  Z,  each  line  is  said 
to  be  ±  to  the  other.) 


Prop.  X.]  PRELIMINARY  PROPOSITIONS.  10 

Proposition  X. 

44.  Theorem.  The  bisectors  of  the  four  angles  which  two 
intersecting  lines  make  with  each  other  form  two  straight 
lines. 

B 


Given  XX'  intersecting  YY'  at  O,  OA  bisecting  Z  XOY, 

OB  bisecting    AYOX'.    00   bisecting    AX'OY' 
and   OD  bisecting  Z  Y'OX. 

To  prove    that  CO  A  and  DOB  are  straight  lines. 

Proof.    1.   A  A  OB  and  BOO  are  rt.  A.  Prop.  IX 

(State  prop.  IX.) 

2.  .'.the  two  together  form  a  st.  angle.  Def.  it.  Z 

(§  19.    State  the  definition.) 

3.  .*.  00 A  is  a  st.  line.  Def.  st.  Z 

(§  14.    State  the  definition.) 

4.  Similarl v  for  DOB. 


45.  The  nature  of  a  logical  proof  should  now  be  understood. 
Before  continuing,  however,  the  following  points  should  be 
emphasized : 

a.  Every  statement  in  a  proof  must  be  based  upon  a  postu- 
late, an  axiom,  a  definition,  or  some  proposition  previously 
considered  of  which  the  student  is  prepared  to  give  the  proof 
again  when  he  refers  to  it. 


20  PLANE    GEOMETRY.  [Intr. 

b.  No  statement  is  true  simply  because  it  appears  to  be  true 
from  a  figure  which  the  student  may  have  drawn,  no  matter 
how  carefully.  Many  cases  will  be  found,  for  example,  where 
angles  appear  equal  when  they  are  not  so. 

c.  The  arrangement  of  the  discussion  of  a  theorem  is  as 
follows : 

Given.  Here  is  stated,  with  reference  to  the  figure  which 
accompanies  the  proof,  whatever  is  given  by  the  theorem. 

To  prove.  Here  is  stated  the  exact  conclusion  to  be  de- 
rived from  what  is  given. 

Proof.  Here  are  set  forth,  in  concise  steps,  the  statements 
to  prove  the  conclusion  just  asserted.  If  the  proof  is  written 
on  the  blackboard,  the  steps  should  be  numbered  for  convenient 
reference  by  class  and  teacher.  The  teacher  will  state  how 
much  in  the  way  of  written  or  indicated  authorities  shall  be 
required  after  each  step. 

Corollarv.  A  corollary  is  a  proposition  so  connected  with 
another  as  not  to  require  separate  treatment.  The  proof  is 
usually  simple,  but  it  must  be  given  with  the  same  accuracy 
as  that  of  the  proposition  to  which  it  is  attached.  It  is  usually 
sufficient  to  say,  This  is  proved  in  step  4 ;  or,  This  follows 
from  steps  2  and  5  by  axiom  3,  etc.  In  every  case  the  stu- 
dent should  (1)  clearly  prove  the  corollary,  but  (2)  do  so  as 
concisely  as  possible.  A  corollary  may  also  follow  from  a 
definition;  thus,  from  the  definitions  of  Proposition  and  Theo- 
rem the  following  might  be  stated  as  a  corollary :  Every 
theorem  is  a  proposition,  but  not  every  proposition  is  a  theo- 
rem; and  as  a  part  of  our  definition  of  a  Perigon  we  incor- 
porated the  corollary  (the  term  then  being  undefined)  that  a 
perigon  equals  two  straight  angles. 

Note.    Any  item  of  interest  may  be  inserted  under  this  head. 


Exercises.  13.    Of  the  proofs  of  the  preliminary  theorems,  state  which 
are  direct  and  which  indirect.     (See  note  on  p.  14.) 

14.    How  can  you  form  a  right  angle  by  paper  folding  ?     Prove  it. 


BOOK    I.  — RECTILINEAR    FIGURES. 


1.    TRIANGLES. 

46.  A  figure  is  any  combination  of  lines  and  points  formed 
under  given  conditions. 

E.g.  an  angle  is  a  figure,  for  it  is  a  combination  of  two  lines  and  one 
point  formed  under  the  condition  that  the  two  lines  proceed  from  the  point. 

47.  A  rectilinear  figure  is  a  figure  of  which  all  the  lines  are 
straight. 

Plane  geometry  treats  of  figures  in  one  plane,  — plane  figures. 

Hence  in  plane  geometry,  which  in  this  work  extends  through  Books 
I  to  V  inclusive,  the  word  figure  used  alone  denotes  a  plane  figure,  and 
all  propositions  and  definitions  refer  to  such  figures  placed  in  one  plane. 

48.  If  the  two  end-points  of  a  broken  line  coincide,  the  fig- 
ure obtained  is  called  a  polygon,  and  the  broken  line  its  perim- 
eter. The  vertices  of  the  angles  made  by  the  segments  of  the 
perimeter  are  called  the  vertices  of  the  polygon,  and  the  seg- 
ments between  the  vertices  are  called  the  sides  of  the  polygon. 

49.  The  perimeter  of  a  polygon  divides  the  plane  into  two 
parts,  one  finite  (the  part  inclosed), 
the  other  infinite.       The  finite  part 
is  called  the  surface  of  the  polygon. 
or  for  brevity  simply  the  polygon. 

A  point  is  said  to  be  within  or 
without  the  polygon  according  as  it 
lies  within  or  without  this  finite  part.  A  polygon. 

The  figure  ABCBE  is  a  polygon  (the  sides  being  produced  for  a  sub- 
sequent definition). 

21 


22 


PLANE    GEOMETRY. 


[Bk. 


50.  In  passing  counter-clockwise  around  the  perimeter  of  a 
polygon  the  angles  on  the  left  are  called  the  interior  angles  of 
the  polygon,  or  for  brevity  simply  the  angles  of  the  polygon. 

Such  are  the  angles  CBA,  DCB,  EDC,  in  the  figure  on  p.  21. 

51.  If  the  sides  of  a  polygon  are  produced  in  the  same  order, 
the  angles  between  the  sides  produced  and  the  following  sides 
are  called  the  exterior  angles  of  the  polygon. 

Such  are  the  angles  XBC,  YCD,  in  the  figure  on  p.  21.     They 

are  the  angles  through  which  one  would  turn,  at  the  successive  corners, 
in  walking  around  the  polygon. 

52.  A  line  joining  the  vertices  of  any  two  angles  of  a  poly- 
gon which  have  not  a  common  arm,  is  called  a  diagonal. 

Such  a  line  would  be  the  one  joining  A  and  C  in  the  figure  on  p.  21. 
The  sides,  angles,  and  diagonals  of  a  polygon  are  often  called  its  paints. 


53.  A  polygon  which  has 
all  of  its  sides  equal  is  called 
equilateral. 


54.  Two  polygons  are  said 
to  be  mutually  equilateral,  or 
one  is  said  to  be  equilateral 
to  the  other,  when  the  sides  of 
the  one  are  respectively  equal 
to  the  sides  of  the  other. 


A  polygon  which  has  all  of 
its  angles  equal  is  called  equi- 
angular. 


Two  polygons  are  said  to 
be  mutually  equiangular,  or 
one  is  said  to  be  equiangular 
to  the  other,  when  the  angles  of 
the  one  are  respectively  equal 
to  the  angles  of  the  other. 


55.    A  polygon  of  three  sides  is  called  a  triangle;   one  of 
four  sides,  a  quadrilateral. 


Secs.  56-59.]  TRIANGLES.  23 

56.  Any  side  of  a  polygon  may  be  called  its  base,  the  side 
on  which  the  hgnre  appears  to  stand  being  usually  so  called, 
as  AB  in  the  figure  on  p.  21. 

In  the  case  of  a  triangle,  the  vertex  of  the  angle  opposite 
the  base  is  called  the  vertex  of  the  triangle,  the  angle  itself 
being  called  the  vertical  angle  of  the  triangle,  and  the  other 
two  angles  the  base  angles. 

Thus,  in  the  first  triangle  on  p.  25,  C  is  the  vertex  of  the  triangle,  Z  C 


57.  Two  figures  which  may  be  made  to  coincide  in  all  their 
parts  by  being  placed  one  upon  the  other  are  said  to  be  con- 
gruent. 

For  example,  two  line-segments  may  be  congruent,  or  two  angles,  or 
two  triangles,  etc. 

58.  The  operation  of  placing  one  figure  upon  the  other  so 
that  the  two  shall  coincide  is  called  superposition,  and  the 
figures  are  sometimes  called  superposable  (a  synonym  of  con- 
gruent). 

This  is  illustrated  in  prop.  I. 

Superposition  is  an  imaginary  operation.  It  is  assumed  as 
a  postulate  (§  61)  that  figures  may  be  moved  about  in  space 
with  no  other  change  than  that  of  position.  The  actual  move- 
ment is,  however,  left  for  the  imagination. 

59.  It  will  hereafter  be  explained  and  defined  that  polygons 
of  the  same  shape  are  called  similar,  the  symbol  of  similarity 
being  ^,  and  that  those  of  the  same  area  are  called  equal  or 
equivalent,  the  symbol  being  =.  Congruent  figures  are  both 
similar  and  equal,  and  hence  the  symbol  for  congruence  is  =, 
a  symbol  used  in  modified  form  by  the  great  mathematician 
Leibnitz. 

The  symbol  ~~  is  derived  from  the  letter  S,  the  initial  of 
the  Latin  si  mills,  similar. 


24 


PLANE    GEOMETRY. 


[Bk.  I. 


Many  writers  use  equal  for  congruent,  and  equivalent  for 
equal,  as  above  defined.  But  because  of  the  various  meanings 
of  the  word   equal,   and   its  general  use   as  a  synonym  for 


Equality. 


Similarity. 


Congruence. 


equivalent,  the  more  exact  word  congruent  with  its  suggestive 
symbol  is  coming  to  be  employed.  The  student  should  be 
familiar  with  this  other  use  of  the  words  equal  and  equivalent. 

60.  It  is  customary  to  designate  the  sides 
of  a  triangle  by  the  small  letters  correspond- 
ing to  the  capital  letters  which  designate  the 
opposite  vertices.  A 

Thus,  in  the  figure,  side  a  is  opposite  vertex  A,  etc. 

61.  It  now  becomes  necessary  to  assume  three  other  pos- 
tulates. 

Postulates  of  Motion. 

1.  A  figure  may  he  moved  about  in  space  with  no  other 
change  than  that  of  position,  and  so  that  any  one  of  its  points 
may  be  made  to  coincide  with  any  assigned  point  in  space. 

That  is,  we  may  pick  up  one  polygon  and  place  it  on  another  without 
changing  its  shape  or  size. 

2.  A  figure  may  be  moved  about  in  space  while  one  of  its 
points  remains  fixed. 

Such  movement  is  called  rotation  about  a  center,  the  center  being  the 
fixed  point. 

3.  A  figure  may  be  moved  about  in  space  while  two  of  its 
points  remain  fixed. 

Such  movement  is  called  rotation  about  an  axis,  the  axis  being  the  line 
determined  by  the  two  points. 


Prop.  I. 


TRIANGLES. 


25 


Proposition  I. 

62.  Theorem.  If  two  triangles  have  two  sides  and  the  in- 
cluded angle  of  the  one  respectively  equal  to  two  sides  and 
the  included  angle  of  the  other,  the  triangles  are  congruent. 


Given 


A  c  b  A' 

the  A  ABC,  A'B'C  such  that 
c  =  c', 
b  =  b',  and 
ZA  =  ZA'. 
To  prove     that  A  ABC  ^  A  A'B'C. 

Proof.    1.  Place  A  A'B'C  on  A  ABC  so  that 
A'  falls  on  A,  and 
c'  coincides  with  its  eqnal  c. 

2.  Then  b'  may  be  caused  to  fall  on  b, 
because  Z  A'  =  Z  A. 

3.  Then  C  will  fall  at  C, 
because  b'  =  b. 


4. 


will  coincide  with  a, 


§61,1 
§  61,  2 

Given  ;   §  61,  3 

Given;   §  57 
§28,1 


5. 


(Two  points  determine  a  straight  line.) 

AABC  =  AA'B'C,  by  definition  of  congruence. 

§  57 

Notes.     This  is  a  proof  by  superposition. 

The  theorem  may  be  stated,  A  triangle  is  determined  when  two  sides 
and  the  included  angle  are  given. 

In  the  exercises  hereafter  given,  the  proofs  are  to  be  given  in  full ; 
when  a  question  is  asked,  a  proof  of  the  answer  is  to  be  given ;  when  a 
theorem  is  suggested,  it  is  to  be  completely  stated  and  then  proved. 


26 


PLANE    GEOMETRY. 


[Bk. 


Proposition  II. 

63.  Theorem.  If  two  triangles  have  two  angles  and  the 
included  side  of  the  one  respectively  equal  to  two  angles  and 
the  included  side  of  the  other,  the  triangles  are  congruent. 


K 


B 


B 


Given  the  A  ABC  and  A'B'C  such  that 

Z.C  =  ZC, 

ZB  =  Z  B',  and 
a  =  a'. 
To  prove     that  A  ABC  ^  A  A'B'C 

Proof.    1.  Place  A  A'B'C  on  A  ABC  so  that  a'  falls  on  a  and 


Z  C  coincides  with  its  equal  Z  C. 

2.  Then  B'  will  fall  on  B  because    a'  =  a. 

3.  Then  c'  will  fall  on  c  because  Z-B' '  =  /LB. 

4.  .*.  A'  will  coincide  with  A. 

(Two  straight  lines  determine  a  point.) 

5.  .'.  A  ABC  ^  A  A'B'C,  by  definition  of  congruence. 

§  57 
Note.     Prop.  II,  and  prop.  Ill  following,  are  attributed  to  Thales. 


§  61 
Given 
Given 

§  28,  2 


Exercises.  15.  In  the  figure  on  p.  19,  given 
that  OA  bisects  angle  XOY,  and  that  OB  is 
perpendicular  to  OA,  prove  that  OB  bisects 
angle  YOX'. 

16.    Show  that  the  distance  BA  across  a  lake 
may  be  measured  by  setting  up  a  stake  at  0,     A' 
sighting  across  it  to  fix  the  lines  A'B  and  B'A, 
laying  off  OA'  —  OA,  and  OB'  —  OB,  and  then  measuring  B'A' 


Sacs.  64, 65.]  TRIANGLES.  27 

64.  Reciprocal  Theorems.  The  student  will  notice  that  prop- 
ositions I  and  II  have  a  certain  similarity.  Indeed,  if  the 
words  side  and  angle  are  interchanged  in  prop.  I.  it  becomes 
prop.  II,  and  if  interchanged  in  prop.  II  that  becomes  prop.  I. 
Theorems  of  this  kind  are  called  reciprocal.  The  relation  is 
more  clearly  seen  by  resorting  to  parallel  columns. 

Prop.  I.     If  two  triangles  have  Prop.  II.     If  two  triangles  Lave 

two  sides  and  the  included  angle  of  two  angles  and  the  included  side  of 

the  one  respectively  equal  to  two  the  one  respectively  equal  to  two 

sides  and  the  included  angle  of  the  angles  and  the  included  side  of  the 

other,  the  triangles  are  congruent.  other,  the  triangles  are  congruent. 

Moreover,  if  small  letters  and  capitals  are  interchanged  in 
the  proof  of  prop.  I,   the  proof  becomes  that  of  prop.   II. 

65.  The  principle  involved  is  called  the  Principle  of  Reci- 
procity, and  is  extensively  used  in  geometry.  But  the  student 
must  not  suppose  that  because  a  theorem  is  true  its  reciprocal 
theorem  is  also  true ;  in  elementary  geometry,  involving 
measurements,  the  reciprocal  is  often  false.  The  principle 
is,  however,  of  great  value  even  here,  for  it  leads  the  student 
to  see  the  relation  between  propositions,  and  it  often  suggests 
new  possible  theorems  for  investigation.  For  these  purposes 
we  shall  use  it. 

At  present  it  is  sufficient  to  say  that  for  many  theorems  of 
plane  geometry  reciprocal  theorems  may  be  formed  by  re- 
placing the  words 

point  by  line, 
line  by  point, 
(ingles  of  a  triangle  by  (opposite^  sides  of  a  triangle, 
sides  of  a  triangle  by  (opposite)  angles  of  a  triangle. 


Exercises.     17.  F.xplain  this  statement  and  tell  why  it  is  true  :  Any  two 

sides  and  the  included  angle  of  a  triangle  determine  the  remaining  parts. 

18.    State  the  reciprocal  of  ex.  17  and  tell  whether  it  is  true,  and  why. 


28  PLANE    GEOMETRY.  [Bk.  I. 

Proposition  III. 

66.    Theorem.     If  two  sides  of  a  trionigle  are  equal,  the 
angles  opposite  those  sides  are  equal. 


Given         the  A  ABC  with  AC  =  BC. 
To  prove     that  AA=Z.B. 

Proof.    1.   Suppose  m  to  bisect  Z.  ba. 

2.  Then  v  b  =  a,  Given 

and  Z.  bin  —  A  ma, 
and  m  =  m, 

3.  . ' .  A  AM C  £  A  BMC,  Prop.  I 

(State  prop.  I.) 
and  AA—/-B,  by  definition  of  congruence.         §  57 

Corollary.  If  a  triangle  is  equilateral,  it  is  also  equi- 
angular. 

For  by  the  theorem  the  angles  opposite  the  equal  sides  are  equal. 

67.  Definitions.  The  line  from  any  vertex  of  a  triangle  to 
the  mid-point  of  the  opposite  side  is  called  the  median  to  that 
side. 

In  the  above  figure,  CM  is  the  median  to  AB. 

If  a  triangle  has  two  equal  sides,  it  is  called  an  isosceles 
triangle. 


Sec.  68.]  TRIANGLES.  29 

The  third  side  is  called  the  base  of  the  isosceles  triangle, 
and  the  equal  sides  are  called  the  sides. 

A  triangle  which  has  no  two  sides  equal  is  called  a  scalene 
triangle. 

The  distance  from  one  point  to  another  is  the  length  of  the 
straight  line-segment  joining  them. 

The  distance  from  a  point  to  a  line  is  the  length  of  the  per- 
pendicular from  that  point  to  that  line. 

That  this  perpendicular  is  unique  will  be  proved  later. 

This  is  the  meaning  of  the  word  distance  in  plane  geometry.  In 
speaking  of  points  on  a  curved  surface  (for  example,  the  earth's  surface), 
distance  may  be  measured  on  a  curved  line. 

68.    In  the  figure  of  prop.  Ill, 

A  AMC  ^  A  BMC,  as  proved. 
.\AM=MB, 

and  Z  CMA  =  Z  BMC, 
and  hence  each  is  a  right  angle. 

In  cases  of  this  kind  the  points  A  and  B  are  said  to  be 
symmetric  with  respect  to  an  axis.  Hence,  in  the  figure,  CM  is 
called  an  axis  of  symmetry.     And,  in  general,  two  systems  of 

points,  Ax,  Bx,  C\,  ,  A2,  B2,  C2, ,  are  said  to  be  symmetric 

vitlt  respect  to  an  axis  when  all  lines,  AXA2,  BXB2,  ,  are 

bisected  at  right  angles  by  that  axis. 

Also,  two  figures  are  said  to  be  symmetric  with  respect  to  an 
axis  when  their  systems  of  points  are  symmetric. 

A  single  figure,  like  that  of  prop.  Ill,  is  said  to  be  sym- 
metric with  respect  to  an  axis  when  this  axis  divides  it  into 
two  svmmetric  figures. 


Exercises.  19.  If  four  lines  go  out  from  a  point  making  four  angles 
of  which  the  first  and  third  are  equal,  and  the  second  and  fourth  are 
equal,  prove  that  the  four  lines  form  two  intersecting  straight  lines. 

20.  In  the  figure  on  p.  10,  if  a  line  passes  through  0  and  bisects 
angle  XOA,  prove  that  it  also  bisects  angle  X'OC 


30  PLANE   GEOMETRY.  [Bk.  I. 

Proposition  IV. 

69.   Theorem.     If  two  angles  of  a  triangle  are  equal,  the 
sides  opposite  those  angles  are  equal. 


A  B 

Given  the  A  ABC  with  Z  A  =  Z  B. 

To  prove     that  a  =  b. 

Proof.    1.   Suppose  that  a  =£  b, 

and  that  a  >  b. 

2.  Then  let  BX,  a  part  of  a,  equal  &,  and  join  A  and  A' 

3.  Then  '.'ZB=Z.BAC,  Given 
and                           J£  =  AB, 

.'.AABC^A  BAX.  Why  ? 

4.  .'.the  supposition  leads  to  an  absurdity,  for 

A  ABO  A  BAX,  Ax.  8 

(State  ax.  8.) 
and  . ' .  a~jj>  b. 
In  the  same  way  it  may  be  shown  that  a  <£  b, 
and  . ' .  a  =  b. 

Corollary.     If  a  triangle  is  equiangular,  it  is  a /so  equi- 
lateral.    (Why  ?)  

Exercise.     21.    If  four  points,  A,  B,  C,  D,  are  placed  in  order  on  a 
line,  and  if  AC  =  BD,  prove  that  AB  =  CD. 


Prop.  V. 


TRIANGLES. 


31 


Proposition  V. 

70.  Theorem.  If  any  side  of  a  triangle  is  produced,  the 
exterior  angle  is  greater  than  either  of  the  interior  angles 
not  adjacent  to   it. 

C 


A  B 

Given  the  A  ABC,  with  AB  produced  to  X. 

To  prove     that  Z  XBC  >  Z  C,  and  also  >  Z  BA  C '. 

Proof.    1.   Suppose  BC  bisected  at  31.  AM  drawn  and  produced 
to  P  so  that  3IP  =  AM,  and  BP  drawn. 

Why  ? 
Why  ? 
§  57 
Why  ? 
Why  ? 

Similarly,  by  producing  CB,  bisecting  AB  at  X, 
producing  CX,  etc..  it  can  be  shown  that  an  angle 
equal  to  Z  XBC  is  greater  than  ABAC. 


2. 

Then    ■ 

.•  Z  BMP  =  Z  CJ/J. 
r.ABPJI^A  CA3I, 

and 

ZPB3T=ZC. 

3. 

But 

Z.XBO  Z.PB3L 

4. 

..Z.XBOZ  C 

Exercises.     22.    Show  that,  in  the  figure  of  prop.  V,  Z  XBC  >  Z  1L4  C 
by  following  out  in  full  the  proof  suggested  in  step  5.  • 

23.  In  the  figure  of  prop.  V,  join  C  to  any  point  in  the  segment  A  B 
and  prove  that  Z  CBA  +  ZBAC<  180°. 

24.  If  a  diagonal  of  a  quadrilateral  bisects  two  angles,  the  quadrilateral 
has  two  pairs  of  equal  sides. 

25.  How  many  equal  lines  can  be  drawn  from  a  given  point  to  a  given 
line  ?     Show  that  if  another  is  supposed  to  be  drawn,  an  absurdity  results. 


32 


PLANE    GEOMETRY. 


[Bk.  I. 


Proposition  VI. 

71.  Theorem.  If  two  sides  of  a  triangle  are  unequal,  the 
opposite  angles  are  unequal  and  the  greater  side  has  the 
greater  angle  opposite. 


Given  the  A  ABC,  with  a  >  b. 

To  prove     that  ZA>ZB. 

Proof.    1.  Suppose  Z  C  bisected  by  YY'  cutting  AB  at  D,  CA' 
made  equal  to  CA,  and  DA'  drawn. 

2.  Then  A. ID  C  ^  A. 4  'DC,  and  ZA  =  Z  CA  'D.  Why? 

3.  But  Z  CA'D  >  AB.  Prop.  V 

(§  70.    If  any  side  of  a  A  is  produced,  the  exterior  angle  is  greater  than 
either  of  the  int.  A  not  adjacent  to  it.) 

4.  ..ZA>ZB.  Subst.  2  in  3 


Exercises.     26.    State,  without  proof,  the  reciprocal  of  prop.  VI. 

27.  Can  a  scalene  triangle  have  two,  equal  angles  ?     Proof. 

28.  Prove  prop.  VI  by  drawing  AA'  instead  of  DA',  and  proving  that 
Z  A  >  Z  A' AC  =  Z  CA'A  >  Z  B. 

29.  ABCD  is  a  quadrilateral  of  which  DA  is  the  longest  side  and  BC 
the  shortest.  Which  is  greater,  ZB  or  ZD?  Prove  it.  (Suggestion  : 
Draw  BD.)     Also  Z  C  or  Z  A  ?     Prove  it. 

30.  How  many  perpendiculars  can  be  drawn  to  a  given  line  from  a 
point  outside  that  line?  Show  that  any  other  supposition  violates 
prop.  V. 

31.  ABC  is  a  triangle  having  Z  B  =  twice  ZA;  Z  B  is  bisected  by  a 
line  meeting  b  at  D ;  prove  that  AD  —  BD. 


Prop.  VII.] 


TRIANGLES. 


33 


Proposition  VII. 


72.  Theorem.  If  two  angles  of  a  triangle  are  unequal, 
the  opposite  sides  are  unequal  and  the  greater  angle  has  the 
greater  side  opposite. 


Given  the  A  ABC  with  Z  A  >  Z  B. 

To  prove     that  a  >  b. 

Proof.    1.  a  4-  b,  for  if  a  =  b,  then  Z  A  =  Z  B.        Why  ? 

2.  a  <  b,  for  if  a  <  b,  then  Z  .i<  Z  J5. 

Prop.  VI.     State  it. 

3.  .'.a  must  be  greater  than  b. 

Note.  It  must  not  be  inferred  from  props.  VI,  VII  that,  because  one 
angle  of  a  triangle  is  twice  as  large  as  another,  one  side  is  twice  as  long 
as  another. 


Exercises.  32.  Prove  that  if  the  bisector  of  any  angle  of  a  triangle  is 
perpendicular  to  the  opposite  side,  the  triangle  is  isosceles. 

33.  Suppose  any  point  taken  on  the  perpendicular  bisector  of  a  line  ; 
is  it  equally  or  unequally  distant  from  the  ends  of  the  line  ?  Give  the 
proof  in  full. 


34  a.  Prove  that  in  an  isosceles 
triangle  ABC,  where  a  =b,  the 
bisector  of  Z  C,  produced  to  c, 
bisects  side  c. 


34  b.  Prove  that  in  an  isosceles 
triangle  abc,  where  ZA  =  ZB,  the 
bisector  of  side  c,  joined  to  C, 
bisects  Z  C. 


35.  After  reading  §  73,  state  the  converse  of  each  of  the  following : 
(a)  prop.  Ill  ;  (b)  prop.  IV  ;  (c)  prop.  VI ;  (d)  prop.  VII  ;  (e)  this  state- 
ment, If  the  animal  is  a  horse,  then  the  animal  has  two  eyes.  Of  these 
converses,  how  many  are  true  ? 

36.  What  kind  of  a  triangle  is  formed  by  joining  the  mid-points  of  the 
sides  of  an  equilateral  triangle  ?     Prove  it. 


34  PL  A  NE   GE  OME  TR  Y.  [B  k  .  I. 

73.  The  Law  of  Converse.  Two  theorems  are  said  to  be  the 
converse,  each  of  the  other,  when  what  is  given  in  the  one  is 
what  is  to  be  proved  in  the  other,  and  vice  versa. 

E.g.  props.  VI  and  VII.  The  converse  of  a  theorem  must  not  be  con- 
fused with  its  reciprocal.    Props.  I  and  II  are  reciprocal,  but  not  converse. 

Because  a  theorem  is  true  its  converse  is  not  necessarily  true. 

For  example,  prel.  prop.  I  may  be  stated  thus  :  Given  that  A  r  and  r/ 
are  rt.  A,  to  prove  that  Z r  =  Z  r' ;  the  converse  is,  Given  that  Zr  =  Zr', 
to  prove  that  they  are  rt.  A.  This  converse  is  evidently  false,  for  Z  r 
could  equal  Zr'  without  their  being  rt.  A. 

But  there  is  one  important  class  of  converse  theorems,  illus- 
trated by  props.  IV  and  VII,  that  should  be  mentioned.  When- 
ever three  theorems  have  the  following  relations,  their  converses 
must  be  true  : 

1.  If  it  has  been  proved  that  when  A>  B,  then  X  >  Y,  and 

2.  "  "  "  A  =  B,     "    X  =  T,    " 

3.  "  "  "  A  <  B,     «    X  <  Y, 
then  the  converse  of  each  of  these  is  true.     For 

1'.  If  X  >  Y,  then  A  can  neither  be  equal  to  nor  less  than 
B,  without  violating  2  or  3  ;  ..  A  >  B.     (Converse  of  1.) 

2'.  If  A'  =  Y,  then  A  can  neither  be  greater  nor  less  than  B, 
without  violating  1  or  3 ;  .'.  A  =  B.     (Converse  of  2.) 

3'.  If  X  <  Y,  then  A  can  neither  be  greater  than  nor  equal  to 
B,  without  violating  1  or  2 ;  .'.  A  <  B.     (Converse  of  3.) 

The  law  just  proved  will  hereafter  be  referred  to  as  the  Law 
of  Converse.  By  its  use  the  proof  of  the  converse  of  many 
theorems,  where  true,  is  made  very  simple. 

The  student  should  not  proceed  further  unless  the  Law  of 
Converse  is  thoroughly  understood,  and  its  proof  mastered. 

Prop.  VII  may  now  be  proved  by  the  Law  of  Converse, 
thus  : 

If  a  >  b,  then  Z  A  >  Z  B.  Prop.  VI 

If  a  =  by     "     Z  A  =  Z  B.  "     III 

Ifa<b,     "     ZA<ZB.  "      VI 

.  * .  eaeh  converse  is  true,  and  if  Z  A  >  Z  B,  then  a  >  b. 


Sec.  74.]  TRIANGLES.  35 

74.  Suggestions  as  to  the  Treatment  of  the  Exercises.  Tims 
far  the  student  has  been  left  to  his  own  ingenuity  in  treating 
the  exercises.     A  few  suggestions  should  now  be  given. 

1.  In  attacking  a.  theorem  take  the  most  general figure  possible. 

E.g.  if  a  theorem  relates  to  a  triangle,  draw  a  scalene  triangle  ;  an 
equilateral  or  an  isosceles  triangle  often  deceives  the  eye,  and  leads 
away  from  the  demonstration.  Draw  all  figures  accurately  ;  an  accurate 
figure  often  suggests  the  demonstration.  But  the  student  who  relies 
too  much  upon  the  accuracy  of  the  figure  in  the  demonstration  itself  is 
liable  to  go  astray. 

2.  Be  certain  that  what  is  given  and  what  is  to  be  proved 
arc  clearly  stated,  with  reference  to  the  letters  of  the  figure. 

This  has  been  done  in  all  of  the  theorems  thus  far  proved.  The  neglect 
to  do  so  in  the  exercises  is  one  of  the  most  fruitful  sources  of  failure. 

3.  Then  begin  by  assuming  the  theorem  true  .-  see  what  fol- 
lows from  that  assumption  •  then  see  if  this  can  be  proved  true 
without  the  assumption  ;    if  so,  try  to  reverse  the  process. 

E.g.  suppose  PO  _L  X'X,  and  PB,  PA  two  obliques  cutting  off  OA  >  OB, 
as  in  the  figure,  and  that  it  is  required  to  prove 
PA  >  PB.  Assume  it  true  ;  then  Z  b  >  Z  a. 
Now  see  if  Zb>  Z  a  without  the  assumption  ; 
Zb>Zc,  which  =  Z  d.  which  > Z a,  by  prop.  V  ; 
.-.  Zb>  Z  a,  without  the  assumption.  Xow  re-  *  o  B  A 
verse  the  process  ;  v  Z  6  >  Za,  .-.  PA  >  PB 
by  prop.  VII. 

4.  Or  begin  by  assuming  the  theorem  false,  and  endeavor  to 
show  the  absurdity  of  the  assumption.    \  lied  actio  ad  absurdum.) 

5.  To  secure  a  clearer  understanding  of  the  theorem  it  is 
often  well  to  follow  Pascal's  advice  and  substitute  the  defini- 
tion for  the  name  of  the  thing  defined. 

E.g.  suppose  it  is  to  be  proved  that  the  median  to  the  base  of  an  isos- 
celes triangle  is  perpendicular  to  the  base.     Instead  of  saying  : 

'•  Given  CAT  the  median  to  the  base  of  the  isosceles  triangle  ABC"  (see 
figure  on  p.  28),  it  is  often  better  to  say  : 

"Given  A  ABC,  with  AC  =  B<7,  and  M  taken  on  AB  so  that  AM 
-  MB,''  for  then  the  facts  stand  out  prominently  without  any  confusing 
terms. 


36 


PLANE   GEOMETRY. 


[Bk.  I. 


Proposition  VIII. 

75.    Theorem.      The  sum  of  any  two  sides  of  a  triangle  is 
greater  than  the  third  side. 

X 


Given         the  A  ABC. 

To  prove    that  a  +  b  >  c. 

Proof.    1.  Suppose         Z  C  bisected  by  CD. 

Then  Z  CD  A  >ZDCB.     Prop.  V.     State  it 

2.  And  • . '  Z  A  CD  =  ZDCB,  Step  1 

.•.ZCfDi>ZiCri). 

.*.  o  >  AD.         Prop.  VII.    State  it 
Similarly,  a  >  DB. 

3.  .'.a  +  b>c. 

Corollary.      The  difference  of  any  two  sides  of  a  triangle 
is  less  than  the  third  side. 

For  if  a  +  b  >  c,  and  c  >  6,  then  a  >  c  —  &,  by  ax.  5. 


Exercises.  37.  Two  equal  lines,  ^4C  and  AD,  are  drawn  on  oppo- 
site sides  of  a  line  AB  and  making  equal  angles  with  it ;  BC  and  BD  are 
drawn.     Show  that  BC  and  BD  also  make  equal  angles  with  AB. 

38.  P,  Q,  E  are  points  on  the  sides  AB,  BC,  CA,  respectively,  of  an 
equilateral  triangle  ABC,  such  that  AP  =  i?Q  =  Ctf  ;  joining  P,  Q,  and 
R,  prove  that  A  PQE  is  equilateral.  (Notice  that  ex.  36  is  merely  a 
special  case  of  this  one.) 


39  a.  The  bisectors  of  the  equal 
angles  of  an  isosceles  triangle  form, 
with  the  base,  an  isosceles  triangle. 


39  b.  The  mid-points  of  the  equal 
sides  of  an  isosceles  triangle  form, 
with  the  vertex,  the  vertices  of  an 
isosceles  triangle. 


Peop.  IX. 


TRIANGLES. 


37 


Proposition  IX. 

76.  Theorem.  If  from  the  ends  of  a  side  of  a  triangle  two 
lines  are  drawn  to  a  point  within  the  triangle,  their  sum  is 
less  than  the  sum  of  the  other  two  sides  of  the  triangle,  hut 
they  contain  a  greater  angle. 

C 


Given  the  A  ABC,   P   a   point  within,  and  BP  and  PA 

drawn. 

To  prove     that  (1)  BP  +  PA  <  a  +  b,  (2)  Z  APB  >  Z  C. 

Proof.    1.  Produce  AP  to  meet  a  at  X. 

Then 

XP  +  PA  =  XA  <  XC  +  b,    Ax.  8  ;   prop.  VIII 
(State  ax.  8  and  prop.  VIII.) 

and  BP  <  BX  +  XP.  Prop.  VIII 

2.  .-.BP  +  XP  +  PA  <  BX  +  XC  +  XP  +  b. 

3.  .'.  BP  +  PA  <  a          +  h. 

which  proves  (1).  Why  ? 

4.  Also, 

Z  APB  >  Z  PXB  >  Z  C  which  proves  (2).  Why  ? 


Exercises.     40  a.    If    the   equal  40  6.    If  the  equal  angles  of  an 

sides  of  an   isosceles  triangle   are  isosceles  triangle  are  bisected,  the 

bisected,  the  lines  joining  the  points  angles  formed  by  the  lines  of  bi- 

of  bisection  with  the  vertices  of  the  section   and    the   equal    sides  are 

equal  angles  are  equal.  equal. 

41.  The  perimeter  of  a  quadrilateral  is  less  than  twice  the  sum  of  its 
two  diagonals. 


38 


PLANE   GEOMETRY. 


[Bk.  I. 


Proposition  X. 

77.  Theorem.  If  two  triangles  have  two  sides  of  the  one 
respectively  equal  to  two  sides  of  the  other,  but  the  included 
angles  unequal,  then  the  third  sides  are  unequal,  the  greater 
side  being  opposite  the  greater  angle. 


Fig.  2.  Fig.  3. 

Given  the  A  AXBXCX  and  A2B2C2,  with  ax  =  a2,  bx  =  b2,  but 

Z.CX>Z.  C2. 
To  prove    that  cx  >  c2. 

Proof.    1.   Suppose  A  A2B2C2  placed  on  AAXBXCX  so  that  b2 
and  bx,  being  equal,  coincide.  §  61 

Then  ' .'  Z  C\  >  Z  C2,  side  a2  must  fall  within  Z  Cx, 
as  in  Fig.  3. 

2.  Suppose   CM  drawn    bisecting   Z  B2  CBX,  and  B2M 
drawn. 

3.  Then  in  A  BXMC,  B23IC, 

CBX  =  CB2,  Given 

CM  =  CM, 
Z  MCBX  =  Z  B2CM.  Step  2 

4.  .\ABXMC^AB2MC, 

and  MB,  =  MB2.  Prop.  I 

5.  But         AM  +  MB2  >  All,.  Prop.  VIII 

.'.  AM '  +  MBX  >  AB2, 
or  cx  >  c2. 

The  proof  is  the  same  when  B2  falls  above  AXBX. 


Prop.  XL]  TRIANGLES.  39 

Proposition  XL 

78.  Theorem.  If  two  triangles  have  two  sides  of  the  one 
respectively  equal  to  two  sides  of  the  other,  but  the  third  sides 
unequal,  then  the  included  angles  are  unequal,  the  greater 
angle  being  opposite  the  greater  third  side. 

Given  A  AlBlCl  and  A2B2C2.  with  a,  =  </,.  hx  =  b2,  c\  >  c2. 

To  prove    that  Z.  C\  >  Z  C2. 

Proof.    1.  It  lias  been  shown  that  if  ax  =  a2,  bx  =  b2, 

and  if  Z  C,  >  Z  6*,,  then  c,  >  c2.  Prop.  X 

2.  And  if  Z  C\  =      «        «      «  =  "  Prop.   I 

3.  «     "  Z  d  <      "        «      «  <  "  Prop.  X 

4.  .'.  the  converses  are  true,  which  proves  the  theorem. 

§  73.    Law  of  Converse 
(Explain  the  Law  of  Converse.     Since  this  law  is  so  often  used, 
it  should  be  reviewed  frequently.) 


Exercises.     42.    Are  props.  X  and  XI  reciprocals  ?   converses  ? 

43.  In  A  ABC,  suppose  CA  >  AB,  and  that  points  P,  Q  are  taken  on 
AB,  CA  respectively,  so  that  PB  =  CQ.     Prove  that  BQ  <  CP. 

44.  Investigate  ex.  43  when  P  is  taken  on  AB  produced,  and  Q  on  AC 
prod Heed. 

45.  The  equal  sides,  AC,  BC,  of  an  isosceles  triangle  ABC  are  pro- 
duced through  the  vertex  to  P  and  Q  respectively,  so  that  AP  =  BQ. 
Prove  that  BP  =  AQ. 

46.  Prove  that  the  straight  line  joining  any  two  points  is  less  than  any 
broken  line  joining  them. 

47.  Prove  that  the  perimeter  of  a  triangle  is  less  than  twice  the  sum 
of  the  three  medians. 

48.  In  a  quadrilateral,  prove  that  the  sum  of  either  pair  of  opposite 
sides  is  less  than  the  sum  of  its  two  diagonals. 

49.  If  the  perpendicular  from  any  vertex  of  a  triangle  to  the  opposite 
side  divides  that  side  into  two  segments,  how  does  each  of  these  segments 
compare  in  length  with  its  adjacent  side  of  the  triangle  ?     Prove  it. 


40  PLANE   GEOMETRY.  [Bk.  1. 

Proposition  XII. 

79.  Theorem.  If  two  triangles  have  the  three  sides  of  the 
one  respectively  equal  to  the  three  sides  of  the  other,  the  tri- 
angles are  congruent. 

C 


Given  A  ABC,  AB'C,  with  AB  =  AB',  BC  =  B'C,  and 
AC  =  AC. 

To  prove    that  A  ABC  ^  A  AB'C. 

Proof.  1.  Suppose  no  side  longer  than  AC.  Then  the  A, 
being  mutually  equilateral,  may  be  placed  with  AC 
in  common,  and  on  opposite  sides  of  AC.    Draw  BB'. 

2.  Then  Z  CBB'  =  Z  BB'C,  Prop.  Ill 
and               Z  B'BA  =  Z  AB'B.  Why  ? 

3.  .-.  Z  CBA  =  Z  AB'C.  Why  ? 

4.  •       .'.A  ABC  ^  A  AB'C.  Why? 

^4C  is  evidently  an  axis  of  symmetry  (§  6S)  in  the 
figure. 

Exercises.  50.  Suppose  three  sticks  to  be  hinged  together  to  form 
a  triangle,  could  the  sides  be  moved  so  as  to  change  the  angles  ?  On 
what  theorem  does  the  answer  depend  ?  How  would  it  be  with  a  hinged 
quadrilateral  ? 

51.  Ascertain  and  prove  whether  or  not  a  quadrilateral  is  determined 
when  the  four  sides  and  either  diagonal  are  given  in  fixed  order. 

52.  Also  when  the  four  sides  and  one  angle  are  given  in  fixed  order. 

53.  How  many  braces  would  it  take  to  stiffen  a  three-sided  plane 
figure  ?  four-sided  ?  five-sided  ? 


Prop.  XIII.] 


TRIANGLES. 


41 


Proposition  XIII. 

80.  Theorem.  If  two  triangles  have  two  angles  of  the  one 
respectively  equal  to  tivo  angles  of  the  other,  and  the  sides 
opposite  one  pair  of  equal  angles  equal,  the  triangles  are 
congruent. 


X  B    Y 


Given         A  AB  C,  A'B' C,  with  Z  A  =  Z  A',  Z  B  =  Z  B\  b  =  b'. 

To  prove    that  A  ABC^  A  A'B'C. 

Proof.    1.  Place  A  A'B'C  on  A  ABC  so  that  A'  falls  at  A, 
A'B'  lies  along  AB,  and  C  and  C  both  lie  on  the 


same  side  of  AB. 


61 


2.  Then  because  Z  A  =  Z  A',  and  b  =  b',  b'  coincides 
with  b,  and  C '  with  C. 

3.  Now  B'  cannot  fall  between  A  and  B,  as  at  X,  for 
then  Z  CXJ,  which  =  Z  B',  would  be  greater  than 
Z  B.  Prop.  V.     State  it 

4.  Neither  can  i?'  fall  on  AB  produced,  as  at  Y,  for  then 
Z  Y,  which  =  Z  I?',  would  be  less  than  Z  B. 

Prop.  V 

5.  .*.  B'  must  fall  at  2?,  and  the  A  are  congruent.    §  57 


Exercises.  54.  If  YO  meets  X'X  at  O,  and  YA,  YB  are  drawn  meet- 
ing X'X dXA,B;  and  if  YA  =  YB,  and  AO^OB,  which  is  the  greater, 
A  A  YO  or  Z  0  Y7i  ? 

55.  Consider  the  diagonals  of  an  equilateral  quadrilateral,  (a)  as  to 
their  bisecting  each  other,  (6)  as  to  the  kind  of  angles  they  make  with 
each  other.     State  the  theorems  which  you  discover  and  prove  them. 


42  PLANE   GEOMETRY.  [Bk.  I. 

Proposition  XIV. 

81.  Theorem.  If  tivo  triangles  have  tivo  sides  of  the  one 
respectively  equal  to  two  sides  of  the  other,  and  the  angles 
opposite  one  pair  of  equal  sides  equal,  then  the  angles  opposite 
the  other  pair  of  equal  sides  are  either  equal  or  supplemental, 
and  if  equal  the  triangles  are  congruent. 

C'  C' 


A^     ~~X  B         U "BT         X' ^B' 

Given         A  ABC,  A'B'C,  with  a  =  a',b  =  b',  Z  B  =  Z  B'. 
To  prove     that  either  (1)  Z  A  =  Z  A'  and  A  ABC ^  A  A'B'C, 
or  (2) ZA  +  ZA'  =  st.  Z. 

Proof.    1.  Place  A  A'B'C  on  Ai^Cso  that  B'  falls  at  B, 

a'  coincides  with  its  equal  a,  and  A'  and  A  fall  on 
the  same  side  of  a.  §  61 

2.  Then       v  Z.B  =  AB',  B'A'  lies  along  BA. 

3.  Then  either  A'  falls  at  ^4,  the  A  are  congruent  and 
Z^4  =  Z^4';  or  else  A'  falls  at  some  other  point  on 
BA,  as  at  X,  and  A  A'B'C  ^  A  XJ5C. 

4.  But  v   CX=b'  =  b, 

.'.  Z.A  =  Z.  CXA.    Prop.  III.  State  it 

5.  And 

v  Z  CXA  +  Z  J5X<?  =  st.  Z,  §  14,  def.  st.  Z 

.-.  Z,4  +  Z.4'  =  st.  Z. 


Exercises.  56.  In  prop.  XIV,  step  3,  X  may  lie  on  BA  produced,  in 
some  cases.    Draw  the  figure  and  prove. 

57.  In  prop.  XIV  prove  that  ZA  and  A  A'  must  be  equal,  if  (1)  they 
are  of  the  same  species  (i.e.  both  right,  both  acute,  or  both  obtuse);  or 
(2)  angles  B  and  B'  are  both  right  angles  ;  or  (3)  b  <£  a. 


Sec.  82.]         PARALLELS  AND  PARALLELOGRAMS.  43 


2.     PARALLELS   AND   PARALLELOGRAMS. 

82.    Definitions.     If  two  straight  lines  in  the  same  plane  do 
not  meet,  however  far  produced,  they 
are  said  to  be  parallel.  _^ 


E.g.   A   and  B  in   the  annexed  figure.    B —    \ 

The  fact  that  A  is  parallel   to  B   is  indi-  \ 

cated  by  the  symbol  A  II  B. 

A  line  cutting  two  or  more  lines  is  called  a  transversal  of 
those  lines. 

In  the  figure  of   the  parallel  lines,   T  is  a  transversal  of  A  and  B. 
The  adjacent  figure  shows  a  transversal  of 
two  non-parallel  lines.    The  figure  on  p.  46  b/a 

shows  a  transversal  of  three  lines. 


The  angles  formed  by  a  transver- 
sal cutting  two  lines  (parallel  or  not) 
have  received  special  names.     Thus,  in  the  annexed  figure, 

a,  b}  c',  d'  are  called  exterior  angles ; 

a',  b',  c,  d  are  called  interior  angles ; 

a  and  c'  are  called  alternate  angles;  also  b  and  d\  c  and  a\ 

b'  and  d ; 

a  and  a'  are  called  corresponding  angles ;  also  b  and  //.  c  and  c\ 

d  and  d'. 


Exercises.  58.  Angle  A,  of  triangle  ABC,  is  bisected  by  a  line  meet- 
ing BC  at  P.  Which  is  the  longer,  AB  or  BP  ?  Prove  it.  Also  CA 
or  PC  ?     Prove  it. 

59.  State  the  reciprocal  of  prop.  XIII,  and  tell  whether  it  is  true 
without  modification.     In  what  proposition  is  your  statement  proved  ? 

60.  If  a  quadrilateral  has  two  pairs  of  equal  sides,  prove  that  it  must 
have  one  pair  and  may  have  two  pairs  of  equal  angles,  depending  upon 
the  arrangement  of  the  sides. 


44  PLANE   GEOMETRY.  [Bk.  I. 

Proposition  XV. 

83.  Theorem.  If  a  transversal  of  two  lines  makes  a  pair 
of  alternate  angles  equal,  then  (1)  any  angle  is  equal  to  its 
alternate  angle,  (2)  any  angle  is  equal  to  its  corresponding 
angle,  and  (3)  any  two  interior,  or  any  two  exterior,  angles 
on  the  same  side  of  the  transversal  are  supplemental. 


Given         a  transversal  cutting  two  lines,  making  equal  alter- 
nate A  d  and  b'  as  in  the  figure. 

To  prove    that  (1)  /.a  =  Z.c', 

(2)  Z.a  =  Za', 

(3)  Z  b  +  Z  c'  =  st.  Z. 
Proof.    1.          Z  d  +  Z  a  =  st.  Z, 

and  Z  V  +  Z  c'  =  st.  Z.  §  14,  del  st.  zl 

2.  .\Z<2-fZ«  =  Z&'  +  Zc'.  Why? 

3.  .'.Zfl^Zc'  which  proves  (1).         Ax.  (?) 

4.  V  Z  c'  =  Z  a',  .'.Zft  =  Za',  which  proves  (2). 

5.  Now       v  Z  ft'  =  Z  rf,  and  Zd  =  Zb,  Why  ? 

.\Zfi'=Z&.  Why? 

But        vZ£'  +  Zc'  =  st.  Z,  §14,  def.  st.  Z 

.-.  Z5  +  Zc'  =  st.  Z. 

Corollaries.     1.    //*  tapo  corresponding  angles  are  equal, 
the  same  three  conclusions  follow. 

2.    If  two  interior  or  two  exterior  angles  on  the  same  side  of 
the  transversal  are  supplemental,  the  same  conclusions  follow. 


Prop.  XVI.]     PARALLELS  AND  PARALLELOGRAMS.  45 

Proposition  XVI. 

84.    Theorem.     If  a  transversal  of  two  lines  makes  a  pair 
of  alternate  angles  equal,  the  two  lines  are  parallel. 


Given  P. and  P',  two  lines,  cut  by  a  transversal  T,  making 

equal  alternate  A  c  and  a'. 

To  prove    that  P  II  P\ 

Proof.  1.  P,  P'  cannot  meet  towards  P',  for  then  Z  c  would 
be  an  ext.  Z  of  a  A,  and  .'.Ac  would  be  greater 
than  Z  a'.  Prop.  V.     State  it 

2.  P,  P'  cannot  meet  towards  P,  for  then  Z  a'  would 
be  greater  than  Z  c.  Why  ? 

3.  ..  P,  P'  cannot  meet  at  all,  and  P  II  P'.     Def.  parallel 
Similarly  any  other  two  alt.  A  may  be  taken  equal. 

Corollaries.     1.    If  two  corresponding  angles  are    equal, 
the  lines  are  parallel. 

For  then  two  alt.  A  are  equal.     Prop.  XV,  cor.  1,  which  says  —  (?) 

2.  If  two  interior  or  two  exterior  angles  on  the  same  side  of 
the  transversal  are  supplemental,  the  lines  are  parallel. 

For  then  two  alt.  A  are  equal.    Prop,  XV,  cor.  2,  which  says  —  (?) 

3.  Two  lines  perpendicular  to  the  same  line  are  parallel. 
(Why?)  

Exercises.  61.  In  prop.  XVI  would  lines 
bisecting  Z.  a'  and  Z  c  be  parallel  ?     Prove  it. 

62.  Show  that  if  a  draughtsman's  square 
slides  along  a  ruler,  as  in  the  annexed  figure, 
BXCX  II  B2cl  and  AXCX  II  A2C2. 


46  PLANE    GEOMETRY.  [Bk.  I. 

85.  Postulate  of  Parallels.  It  now  becomes  necessary  to 
assume  another  postulate,  and  upon  it  rests  much  of  the  ele- 
mentary theory  of  parallels.  It  is  :  Tivo  intersecting  straight 
lines  cannot  both  be  parallel  to  the  same  straight  line. 

Corollary.  A  line  cutting  one  of  two  parallel  lines  cuts 
the  other  also,  the  lines  being  unlimited. 

(Show  that  the  corollary  is  necessarily  true  if  the  postulate  is.) 


Proposition  XVII. 

86.    Theorem.      The  alternate,  angles  formed  by  a  trans- 
versal with  two  parallels  are  equal. 


Given         P  and  P'  two  parallels,  and  T,  sl  transversal. 

To  prove    that  any  Z  c  equals  its  alternate  Z  a'. 

Proof.    1.   Suppose  Z  c  >  Z  a',  and  that  Q  is  drawn  as  in  the 
figure,  making  an  Z  equal  to  Z.a'. 

2.  Then  Q  would  be  parallel  to  P' .  Why  ? 

3.  But  this  would  be  impossible,  V  P  II  P\  §  85 

(Two  intersecting  straight  lines  cannot  both  be  parallel  to  the  same 
straight  line.) 

4.  Similarly,  it  is  absurd  to  suppose  that  Z  <i'  >  Ac. 

Corollaries.     1.   A  line  perpendicular  to  one  of  two  paral- 
lels is  perpendicular  to  the  other  also. 

For  it  cuts  the  other  (§  85,  cor.)  and  the  alternate  angles  are  equal 
right  angles. 


Prop.  XVII.]     PARALLELS  A XI)  PARALLELOGRAMS.  47 

2.  A  line  cutting  two  parallels  makes  corresponding  angles 
equal,  and  the  interior,  or  the  exterior,  angles  on  the  same  side 
of  the  transversal  supplemental. 

For  the  alternate  angles  are  equal  (prop.  XVII),  and  hence  prop.  XV 
applies. 

3.  If  the  alternate  or  the  corresponding  angles  are  unequal, 
or  if  the  interior  angles  on  the  same  side  of  the  transversal 
are  not  supplemental,  then  the  Jims  are  not  parallel,  but  meet 
on  that  side  of  the  transversal  on  which  the  sum  of  the  inte- 
rim' angles  is  less  than  a  straight  angle. 

For  the  lines  cannot  be  parallel,  by  prop.  XVII  and  cor.  2. 

Further,  suppose  Z  c  +  Z  b'  <  st.  Z  ; 

then  v  Za'-f //)'=  st.  Z, 

it  follows  that  Zc  <  Za'. 

.-.  P  and  P/  cannot  meet  towards  P\  for  then  Z  c  would  be  greater 
than  Z  a',  prop.  V. 

Let  the  student  give  the  proof  in  fidl  furni,  in  steps. 

4.  Two  lines  respectively  perpendicular  to  two  intersecting 
lines  cannot  be  parallel. 

For,  in  the  annexed  figure,  let  AB  _L  X.  CD  _L  Y;     y^ 
join  A  and  C.     Then  Z  EAC  <  rt.  Z,  and  ZACD 
<  rt.  Z  ;  .-.  their  sum  is  <  st.  Z  ;  .-.  cor.  3  applies.  A/~ 

Give  proof  in  full  form  in  steps.  \p       /q 


5.    If  the  arms  of  one  angle  are  parallel  or  perpendicular 
to  the  arms  of  another,  the  angles  are  equal  or  supplemental. 

The  proof  is  left  to  the  student. 


Exercises.  63.  In  the  figure  of  prop.  XV,  suppose  a  =  c'  —  120°  30', 
how  large  is  each  of  the  other  angles  ? 

64.  In  the  same  figure,  suppose  a  +  d'  =  st.  Z,  and  a  —  2  d.  how  large 
is  each  of  the  other  angles  ? 

65.  If  a  transversal  cuts  two  lines  making  the  sum  of  the  two  interior 
angles  on  the  same  side  of  the  transversal  a  straight  angle,  one  of  them 
being  30°  27',  how  large  is  each  of  the  other  angles  ? 


48 


PLANE    GEOMETRY. 


[Bk.  I. 


Proposition   XVIII. 

87.    Theorem.     Lines  parallel  to  the  same  line  are  paral- 
lel to  each  other. 


Given  A  II  M,  and  B  II  M. 

To  prove    that  A  II  B. 

Proof.    1.  Suppose    T  a    transversal,    making    corresponding 
A  a,  m,  b,  with  A,  31,  B,  respectively. 

2.  Then  v  A\\M, 

.-.Aa  =  A  m.         Prop.  XVII,  cor.  2 

3.  And         v  B\\  M,  .\Z.b  =  Z.m. 

4.  .\/.a  =  £b.  .        Why? 

5.  .'.  AWB.     Prop.  XVI,  cor.  1.    State  it 


Exercises.  66.  In  prop.  XVIII,  if  T  cuts  A,  must  it  necessarily  cut 
M  ?     Why  ?     If  it  cuts  M,  must  it  necessarily  cut  B  ?     Why  ? 

67.  Prove  that  a  line  parallel  to  the  base  of  an  isosceles  triangle 
makes  equal  angles  with  the  sides  or  the  sides  produced.  (The  line  may 
pass  above,  through,  or  below  the  triangle,  or  through  the  vertex.) 

68.  If  through  any  point  equidistant  from  two  parallels,  two  transver- 
sals are  drawn,  prove  that  they  will  cut  off  equal  segments  of  the  parallels. 

69.  ABC  is  a  triangle,  and  through  P,  the  point  of  intersection  of 
the  bisectors  of  Z  B  and  ZC,a  line  is  drawn  parallel  to  BC,  meeting  AB 
at  M,  and  CA  at  N.     Prove  that  MN  =  MB  +  CN. 

70.  Through  the  mid-point  of  the  segment  of  a  transversal  cut  off  by 
two  parallels,  a  straight  line  passes,  terminated  by  the  parallels.  Prove 
that  this  line  is  bisected  by  the  transversal. 


Prop.  XIX.]    PARALLELS  AND  PARALLELOGRAMS.  49 

Proposition   XIX. 

88.  Theorem.  In  any  triangle,  (1)  any  exterior  angle  equals 
the  sum  of  the  two  interior  non-adjacent  angles  ;  (2)  the  sum 
of  the  three  interior  angles  is  a  straight  angle. 


A  B 

Given  A  ABC,  with  AB  produced  to  X. 

To  prove    that     (1)  AXBC  =  AA  +  A  C; 

(2)  AA  +  AB  +  AC  =  st.  Z. 
Proof.    1.  Suppose  BY  II  AC,  and  A  named  as  in  the  figure. 

2.  Then  Z  x  =  Z  a,  Why  ? 
and                       Ay  =  Ac.                                  Why  ? 

3.  .'.Ax  +  Ay,  or  Z  XBC,  =  Z  «  +  Ac, 

which  proves  (1).  Ax.  2 

4.  But         Ax  +  Ay  +  Ab  =  st.  A.  Def.  st.  Z 

5.  .'.  Aa  +  Ab  +  Ac  =  st.  Z, 

by  substituting  3  in  4,  which  proves  (2). 

Notes.    1.  Prop.  XIX,  (2),  is  attributed  to  Pythagoras. 
2.  The  theorem  is  one  of  the  most  important  of  geometry.     To  it  and 
to  its  corollaries  (p.  50)  frequent  reference  is  hereafter  made. 


Exercises.  71.  PQR  is  a  triangle  having  PQ  =  PR  ;  RP  is  produced 
to  S  so  that  PS  =  RP;   QS  is  drawn.     Prove  that  QS  _L  RQ. 

72.  Prove  prop.  XIX,  (2),  by  drawing  through  C,  in  the  figure  given, 
aline  II  AB. 

73.  Also  by  assuming  any  point  P  on  AB,  drawing  PC,  and  showing 
that  Z  BPC  +  Z  CPA  =  st.  Z ,  and  also  equals  the  sum  of  the  interior 
angles. 

74.  State  the  reciprocal  of  prop.  VIII,  and  prove  or  disprove  it. 


50  PLANE    GEOMETRY.  [Bk.  1. 

Corollaries  to  prop.  XIX.     1.    If  a  triangle    has   one 
rigid  angle,  or  one  obtuse  angle,  the  other  angles  are  acute. 
For  the  sum  of  all  three  is  a  straight  angle. 

2.  Every  triangle  has  at  least  two  acute  angles.    V 

For  if  it  had  none  or  only  one,  the  sum  of  the  others  would  equal  or 
exceed  what  kind  of  an  angle,  and  thus  violate  what  theorem  ? 

3.  From  a  point  outside  a  given  line  not  more  than  one 
perpendicular  can  be  drawn  to  that  line. 

For  if  two  could  be  drawn,  a  triangle  could  be  formed  having  how 
many  right  angles,  thus  violating  what  corollary  ? 

4.  If  a  triangle  has  a  right  angle,  the  two  acute  angles  are 
complement  al. 

For  the  sum  of  all  three  must  equal  two  right  angles ;  therefore,  etc. 

5.  If  tivo  triangles  have  tivo  sides  of  the  one  respectively 
equal  to  two  sides  of  the  other,  and  the  angles  opposite  one 
pair  of  equal  sides  right  angles,  or  equal  obtuse  angles,  the 
triangles  are  congruent.  I 

For  prop.  XIV  then  applies ;  the  oblique  angles  cannot  be  supplemental. 

6.  If  two  angles  of  one  triangle  equal  two  angles  of  another, 
the  third  angles  are  equal.      (Why  ?) 

7.  Two  triangles  are  congruent  if  two  angles  and  any  side 
of  the  one  are  respectively  equal  to  the  corresponding  ptarts  of 
the  other.      (Why  ?) 

8.  Each  angle  of  an  equilateral  triangle  is  one-third  of  a 
straight  angle.      (Why  ?) 

89.  Definitions.  A  triangle,  one  of  whose  angles  is  a  right 
angle,  is  called  a  right-angled  triangle. 

A  triangle,  one  of  whose  angles  is  an  obtuse  angle,  is  called 
an  obtuse-angled  triangle. 

A  triangle,  all  of  whose  angles  are  acute,  is  called  an  acute- 
angled  triangle. 

The  side  opposite  the  right  angle  of  a  right-angled  triangle 
is  called  the  hypotenuse. 


Sec.  90.]     PARALLELS  AND  PARALLELOGRAMS.  51 

90.    Summary  of  Propositions  concerning  Congruent 

Triangles.  Two  triangles  are  congruent  if  the  following 
parts  of  the  one  are  equal  to  the  corresponding  parts  of  the 
other  : 

1.  Two  sides  and  the  included  angle.  Prop.  I 

2.  Two  angles  and  the  included  side.  Prop.  II 

3.  Three  sides.  Prop.  XII 

4.  Two  angles  and  the  side  opposite  one,  Prop.  XIII 

or,  more  generally,  two  angles  and  a  side. 

Prop.  XIX,  cor.  7 

5.  Two  sides  and  the  angle  opposite  one, provided  that  angle 
Js  not  acute.  Prop.  XIV,  and  prop.  XIX,  cor.  5 

If  the  angle  is  acute,  then  from  two  sides  and  the  acute  angle  opposite 
one  of  them  two  different  triangles  may  be  possible.  This  is  therefore 
known  as  the  ambiguous  case.  If  the  side  opposite  the  acute  angle  is  not 
less  than  the  given  adjacent  side,  the  case  is  not  ambiguous.  Why  ?  Draw 
the  figures  illustrating  the  ambiguous  case. 

These  propositions  can  be  summarized  in  one  general  propo- 
sition :  A  triangle  is  determined 'when  any  three  independent 
parts  are  given,  except  in  the  ambiguous  case. 

It  should  be  noted  that  the  three  angles  are  not  three  independent  parts, 
since  when  any  two  of  them  are  given  the  third  is  determined.  (Prop. 
XIX.)  

Exercises.  75.  In  a  right-angled  triangle,  the  mid-point  of  the  hypote- 
nuse is  equidistant  from  the  three  vertices.  (Suppose  a  line  drawn  from 
the  vertex  C  of  the  right  angle  making  with  a  an  angle  equal  to  Z  B.) 

76.  In  a  right-angled  triangle,  a  perpendicular  let  fall  from  the  vertex 
of  the  right  angle,  upon  the  hypotenuse,  cuts  off  two  triangles  mutually 
equiangular  to  the  original  triangle. 

77.  If  a  JL  x  and  b  ±  y,  and  x  intersects  y,  then  Zab  —  Z  xy. 

78.  In  the  annexed  figure,  Z  aai  =  Z  bbx.     Prove  that 
(1)  Z  aa\  —  Z  ab  +  Z  ba-i  ;  (2)  Z  bbx  =  Z  6a i  +  Z  ai&i.  b,l 

79.  How  many  degrees  in  each  angle  of  an  isosceles  /"b 
right-angled  triangle  ?  also  of  an  isosceles  triangle  whose 
vertical  angle  is  72°  ?    178°?   60°? 


52 


PLANE    GEOMETRY. 


[Bk.  I. 


Proposition  XX. 

91.  Theorem.  Of  all  lines  drawn  to  a  given  line  from  a 
given  external  point,  the  perpendicular  is  the  shortest;  of 
others,  those  making  equal  angles  with  the  perpendicular 
are  equal ;  and  of  two  others,  that  which  makes  the  greater 
angle  with  the  perpendicular  is  the  greater. 


Given         PO±XX';  FA,   FA',  FB,  oblique   to   XX1,  with 
Z  A'FO  =  Z  OF  A  <  Z  OFB. 


To  prove 

that 

(1)  PO  <  PA, 

(2)  PA'  =  PA, 

(3)  FB  >  PA,  or  FA'. 

Proof.    1. 

ZPAO<  ZtAOP.     Prop.  XIX,  cor.  1 

2. 

.'.  PO  <  PA,  which  proves  (1). 

Prop.  VII 

3. 

Z  AOP  =  Z  POA',              Prel.  prop.  I 
£A'PO  =  £OPA,                         AVhy? 

and  PO  =  PO. 

4.  .-.  AAOF^AA'OP, 

and  PA'  =  PA,  which  proves  (2).    Why  ? 

5.  Z  BAP  is  obtuse,  v  it  >  Z  AOP,  Prop.  V 
and  Z  PB 0  is  acute,  ' . *  Z  />' OP  is  rt.  Why  ? 

6.  ..  PB>  PA,  or  its  equal  PJ', 

by  step  4,  which  proves  (3).  Prop.  VII 


Prop.  XX.]     PARALLELS  AND  PARALLELOGRAMS.  53 

Corollaries.     1.    From  a  given  external  point  tit  ere  can  be 
two,  and  only  two,  equal  obliques  of  given  length  to  a  given  line. 
Prove  it  by  a  reductio  ad  absurdum. 

2.  If  from  a  point  not  on  a  perpendicular  drawn  t<>  a  line 

at  its  mid-point,  lines  are  drawn  to  the  ends  of  the  line,  these 
lines  are  unequal  and  the  one  cutting  the  perpendicular  is  the 
greater.  , 

Let  Z  be  the  point,  not  on  OP,  in  the  figure.  Suppose  ZA'  to  cut  OP 
at  Y.     Then  ZA'  =  ZY+YA>ZA. 

3.  The  converse  of  cor.  2  is  true. 

For  if  ZA'  cuts  OP,  then  ZA'  >  ZA,  by  cor.  2. 
"  ZA      "      "        "       "    <•  »      '•     "    " 

"  Z      is  on    "        "       "    =   "       (Why  ?) 
.-.  the  Law  of  Converse  (§  73)  evidently  applies  to  this  case. 

4.  Of  two  obliques  from  a  point  to  a  line,  that  which  meets 
the  line  at  the  greater  distance  from  the  foot  of  the  perpen- 
dicular  is  the  greater. 

For  if  OR>OA,  then  Z  OPB  >  Z  OP  A.   (Why  ?)  .-.  prop.  XX  applies. 

5.  Two  obliques  from  a  point  to  a  line,  meeting  that  line  at 
equal  distances  from  the  foot  of  the  perpendicular,  are  equal, 
make  equal  angles  with  this  line  and  also  with  the  perpen- 
dicular. 

Give  the  proof  in  full. 

6.  Two  equal  obliques  from  a  point  to  a  line  cut  off  equal 
segments  from  the  foot  of  the  perpendicular. 

Draw  the  figure.  It  will  then  be  seen  that  prop.  XIX,  cor.  5,  applies. 
The  _L  is  evidently  an  axis  of  symmetry  (§  68). 


Exercises.  80.  -  A  line  perpendicular  to  the  bisector  of  any  angle  of  a 
triangle  makes  an  angle  with  either  arm  of  that  angle  equal  to  half  the 
sum  of  the  other  two  angles  ;  and,  unless  parallel  to  the  base,  it  makes  an 
angle  with  the  line  of  the  base  equal  to  half  the  difference  of  those  angles. 

81.  In  an  isosceles  triangle,  the  perpendicular  from  the  vertex,  the 
median  to  the  base,  and  the  bisector  of  the  vertical  angle  all  coincide. 


54 


PLANE   GEOMETRY. 


[Bk.  I. 


92.    Definitions.     A  polygon  is  said  to  be  convex  when  no 
side  produced  cuts  the  surface  of  the  polygon. 

A  polygon  is  said  to  be  concave  when  a  side  produced  cuts 
the  surface  of  the  polygon. 

A  polygon  is  said  to  be  cross  when  the  perimeter  crosses 
itself. 

The  word  polygon  is  understood,  in  elementary  geometry,  to  refer  to 
a  convex  or  concave  polygon  unless  the  contrary  is  stated. 


Convex. 


Concave. 


Cross. 


A  general  quadrilateral. 


If  all  of  the  sides  of  a  polygon  are  indefinitely  produced, 
the  figure  is  called  a  general 
polygon. 

If  a  polygon  is  both  equi- 
angular and  equilateral,  it  is 
said  to  be  regular. 

By  the  term  regular  polygon,  a 
regular  convex  polygon  is  under- 
stood unless  the  contrary  is  stated. 

A  polygon  is  called  a  tri- 
angle, quadrilateral,  penta- 
gon, hexagon,  heptagon, 
octagon,    nonagon,   decagon, 

dodecagon, pentedecagon, n-gon 

has  3,  4,  5,  6,  7,  8,  9,  10, 12, 15,  ... 

The  student,  even  if  unacquainted  with  Latin  or  Greek,  should  under- 
stand the  derivation  of  these  common  terms.  From  the  Latin  are  derived 
the  words  and  prefix  tri-angle  (three-angle),  quadri-lateral  (four-side), 
nona-  (nine) ;  from  the  Greek  are  derived  poly-gon  (many-angle),  penta- 
(five),  hexa-  (six),  hepta-  (seven),  octo-  (eight),  deca-  (ten),  dodeca-  (twelve), 

Much  light  will  be  thrown  on  the  meaning  of  various  geometric 

terms  by  consulting  the  Table  of  Etymologies  in  the  Appendix. 


Regular  convex 
polygon. 


Regular  cross 
polygon. 

according  as  it 
?i-sides. 


Prop.  XXI.  1     PA RALLELS  A N I)  PA  II A  LL E L  OG  RA  MS.  55 

Proposition  XXI. 

93.    Theorem.      The  sum  of  the  interior  angles  of  an  n-gon 
is  (n  —  2)  straight  angles. 


P 
Given  1\  a  polygon  of  n  sides. 

To  prove    that  the  sum  of  the  interior  angles  is  (u  —  2)  straight 
angles. 

Proof.    1.  P  may  be  divided  into  (n  —  2)  A  by  diagonals  which 
do  not  cross  ;  for, 

(a)  A  4-gon  (quadrilateral)  is  a  A  +  a  A, 

.-.  'J  A,  or  (4-2)  A. 

(b)  A  5-gon  (pentagon)  is  a  4-gon  +  a  A, 

.-.  3  A,  or  (5-2)  A. 

(c)  A  6-gon  (hexagon)  is  a  5-gon  +  a  A, 

.-.  4  A.  or  (6-2)  A. 

(d)  And  every  addition  of  1  side  adds  1  A. 

(e)  .*.  for  an  n-gon  there  are  (n  —  2)  A. 

2.  The  sum  of  the  A  of  each  A  is  a  st.  Z.    Prop.  XIX 

3.  .*.  the  sum  of  the  interior  A  of  an  w-gon  is  (//  —  2) 
st.  A,  because  these  equal  the  sum  of  the  A  of  the  A. 

Corollary.     If  each  of  two  angles  of  a  quadrilateral  is  a 
right  angle,  the  other  two  angles  are  supplemental.     (Why  ?) 


Exercise.     82.    How  many  diagonals  in  a  common  convex  pentagon  ? 
hexagon  ? 


56  PLANE    GEOMETRY.  [Bk.  I. 

94.  Generalization  of  Figures.  If  a  thermometer  registers  70° 
above  zero,  it  is  ordinarily  stated,  in  scientific  works,  that  it 
registers  +  70°,  while  10°  below  zero  is  indicated  by  —  10°, 
the  sign  changing  from  +  to  —  as  the  temperature  decreases 
through  zero.  Similarly,  west  longitude  is  represented  by  the 
sign  -f-,  while  longitude  on  the  other  side  of  0°  (i.e.  east)  is 
represented  by  the  sign  — ,  the  longitude  changing  its  sign  in 
passing  through  zero.  So  in  speaking  of  temperature  it  is 
said  that  10°  +  (—  10°)  =  0,  meaning  thereby  that  if  the  tem- 
perature rises  10°  from  0,  and  then  falls  10°,  the  result  of 
the  two  movements  is  the  original  temperature,  0. 

This  custom  holds  in  geometry.  Thus,  in  this  figure,  if  the 
segment  between  B  and  C  is  thought  of  as  extending  from  B 
to  C,  it  would  be  named  BC ;  and, 

as  is  usually  done  in  geometry  with  , ~ 

lines  thought  of  as  extending  to  the 

right,  it  would  be  considered  Sijyositive  line.  But  if  it  is  thought 
of  as  extending  from  C  to  B,  it  would  be  named  CB,  and  con- 
sidered a  negative  line.  Hence  it  is  said  that  BC  +  CB  =  0, 
an  expression  borrowed  from  algebra,  where  it  would  appear 
in  a  form  like  x  +  (—  x)  =  0. 

Similarly,  with  regard  to  angles :  the  turning  of  an  arm  in 
a  sense  opposite  to  that  of  a  clock-hand,  counter-clockwise, 
is  considered  positive,  while  the  turning 
in  the  opposite  sense  is  considered  nega- 
tive. Thus,  Z  XOA  is  considered  posi- 
tive, but  the  acute  Z  AOX  is  considered 
negative,  and  this  is  indicated  by  the  state- 
ment, —  Z.XOA  —  acute  Z  A  OX.  Hence,  as  in  the  case  of 
lines,  Z  XOA  +  (-  Z  XOA)  =  Z  XOA  +  acute  Z  AOX  =  zero. 
On  this  account  we  pay  special  attention  to  the  manner  of 
lettering  angles,  distinguishing  between  Z  XOA  and  Z  A  OX. 
It  is  only  recently  that  negative  angles  have  been  considered 
in  elementary  geometry,  and  hence  the  older  works  paid  no 
attention  to  the  order  of  the  naming  of  the  arms. 


Sec.  95.]     PARALLELS  AND  PARALLELOGRAMS.  57 

95.  These  considerations  enable  us  to  generalize  man)-  fig- 
ures, with  interesting  results.  Thus,  prop.  XXI  is  true  for  a 
cross  polygon  as  well  as  for  the  simple  cases  usually  consid- 
ered.    If,  in  Fig.  1,  P  is  moved  through  AB  to  the  position 

C 


Fig.  1.  Fig.  2. 


shown  in  Fig.  2,  we  shall  still  have  Z  A  (which  has  passed 
through  0  and  has  become  negative)  +  Zi?-fZC  +  ZP 
(which  is  now  reflex)  =  2  st.  angles. 


Exercises.     83.    Prove  the  last  statement  made  above. 

84.  How  many  points  of  intersection,  at  most,  of  the  sides  of  a  gen- 
eral quadrilateral  ?  pentagon  ? 

85.  How  many  diagonals,  at  most,  has  a  general  quadrilateral  ? 

86.  Prove  prop.  XXI  by  connecting  each  vertex  with  a  point  0  within 
the  figure,  thus  forming  n  &,  giving  n  st.  A,  and  then  subtracting  the  two 
around  0. 

87.  In  an  isosceles  triangle,  the  perpendiculars  from  the  ends  of  the 
base  to  the  opposite  sides  are  equal. 

88.  H  the  bisector  of  the  vertical  angle  of  a  triangle  also  bisects  the 
base,  the  triangle  is  isosceles. 

89.  H  the  base  AB  of  A  ABC  is  produced  to  X,  and  if  the  bisectors 
of  Z  XBC  and  ABAC  meet  at  P,  what  fractional  part  is  A  P  of  A  C  ? 

90.  Given  two  parallels  and  a  transversal,  what  angle  do  the  bisectors 
of  the  interior  angles  on  the  same  side  of  the  transversal  make  with  each 
other  ? 

91.  H  one  angle  of  an  isosceles  triangle  is  given,  and  it  is  known 
whether  it  is  the  vertical  angle  or  not,  then  the  other  two  angles  are 
determined. 


5$ 


PLANE   GEOMETRY. 


[Bk. 


Proposition  XXII. 

96.    Theorem.      The  sum  of  the  exterior  angles  of  any  poly- 
gon is  a  perigon. 


Given         P  and  Q,  two  ^-gons. 

To  prove    that  the  sum  of  the  exterior  A  =  360°  in  each  »-gon. 

Proof.    1.  In  P,   each   interior  Z  -f  its    adjacent    exterior   Z 
=  180°.  §  14,  def.  st.  Z 


2.  /.  sum  of  int.  and  ext.  A—  n  -1800 

3.  But  sum  of  int.  A  =  (n  -  2)  •  180°. 


Ax.  6 

Why  ? 

Ax.  3 


4.     .-.  sum  of  ext.  A  =  2  •  180°  =  360°. 
The  proof  for  Q  is  the  same,  if  Z  a  is  considered  negative. 


Exercises.     92.   Each  exterior  angle  of  an  equilateral  triangle  equals 
how  many  times  each  interior  angle  ? 

93.  Each  exterior  angle  of  a  regular  heptagon  equals  what  fractional 
part  of  each  interior  angle  ? 

94.  Each  exterior   angle  of   a  regular  n-gon  equals  what  fractional 
part  of  each  interior  angle  ?     See  if  the  result  found  is  true  if  n  =  3,  or  4. 

95.  Is  it  possible  for  the  exterior  angle  of  a  regular  polygon  to  be 
70°  ?  72°?  75°?  120°? 

96.  Prove  prop.  XXII  independently  of  prop.  XXI  by  taking  a  point 
anywhere  in  the  plane  of  the  figure  (inside  or  outside 
the  polygon,  or  on  the  perimeter)  and  drawing  par- 
allels to  the  sides  from  that  point,  and  showing  that 
the  sum  of  the  exterior  angles  equals  the  perigon 
about  that  point. 


Sec  97.]  PARALLELS   AND   PARALLELOGRAMS.  59 

97.  Definitions.  A  quadrilateral  whose  opposite  sides  are 
parallel  is  called  a  parallelogram. 

A  quadrilateral  that  lias  one  pair  of  opposite  sides  parallel 
is  called  a  trapezoid. 

Trapezium  is  a  term  often  applied  to  a  quadrilateral  no  two  of  whose 
sides  are  parallel. 

By  the  definition  of  trapezoid  here  given  it  will  be  seen  that  the  paral- 
lelogram may  be  considered  a  special  form  of  the  trapezoid. 

The  parallel  sides  of  a  trapezoid  are  called  its  bases,  and  are  distin- 
guished as  upper  and  lower. 

If  the  two  opposite  non-parallel  sides  of  a  trapezoid  are  equal,  the 
trapezoid  is  said  to  be  isosceles. 

Q      Upper  Base 


c 

Lower  Base 


D^ 

Upper  Base       „ 

.  /         Lower  Base         \  „ 

A 

b 

Parallelogram.  Trapezoid.  Isosceles  trapezoid. 

Iii  the  above  figures,  angles  A,  7>,  or  £,  C,  or  C,  D,  or  D,  A,  are  called 
consecutive  angles.     Angles  A,  C,  or  B,  I),  are  called  opposite  angles. 


V 


Exercises.  97.  If  the  student  has  proved  ex.  96,  let  him  prove  prop. 
XXI  from  that. 

\.    Prove  that  the  quadrilateral  formed  by  the  bisectors  of  the  angles 
of  any  quadrilateral  has  its  opposite  angles  supplemental. 

99.    Show  that  in  ex.  98  the  angles  bisected  may  be  either  the  four 
interior  or  the  four  exterior  angles. 

100.  If  from  the  ends  of  the  base  of  an  isosceles  triangle  perpendicu- 
lars are  drawn  to  the  opposite  sides,  a  new  isosceles  triangle  is  formed, 
each  of  its  base  angles  being  half  the  vertical  angle  of  the  original  triangle. 

101.  The  hypotenuse  is  greater  than  either  of  the  other  sides  of  a  right- 
angled  triangle. 

102.  From  the  vertex  of  the  right  angle  of  a  right-angled  triangle,  is  it 
possible  to  draw,  to  the  hypotenuse,  a  line  longer  than  the  hypotenuse  ? 
Proof. 

103.  A  line  from  the  vertex  of  an  isosceles  triangle  to  any  point  on  the 
base  produced  is  greater  than  either  side.  Is  this  also  true  for  a  scalene 
triangle  ? 


60 


PLANE    GEOMETRY 


[Bk.  I. 


Proposition  XXIII. 

98.  Theorem.  Any  two  consecutive  angles  of  a  parallelo- 
gram are  supplemental,  and  any  two  opposite  angles  are 
equal. 

D 


7 


Given  O  ABCD. 

To  prove    that       .   (1)  Z  A  +  Z  B  =  st.  Z, 

(2)  AA  =  AC. 

Proof.    1.       Z  ^  +  Z  7?  =  st.  Z,  which  proves  (1). 

Prop.  XVII,  cor.  2 

2.  Z  B  +  Z  C  =  st.  Z.  Why  ? 

3.  .\ZA  +  ZB  =  ZB  +  ZC.  Why  ? 

4.  .-.  Z  .4  =  Z  C,  which  proves  (2).  Why  ? 

Corollary.      7/  o«e  angle  of  a  parallelogram  is  a   right 
angle,  all  of  its  angles  are  right  angles.     (Why  ?) 

99.    Definitions.     If  one  angle  of  a  parallelogram  is  a  right 
angle,  the  parallelogram  is  called  a  rectangle. 

By  the  corollary,  all  angles  of  a  rectangle  are  right  angles. 

A  parallelogram  that  has  two  adjacent  sides  equal  is  called 
a  rhombus. 

It  is  shown  in  prop.  XXIV,  cor.  1,  that  all 
of  its  sides  are  equal. 


A   rectangle   that    has   two   adjacent 
sides  equal  is  called  a  square. 

It  is  shown  in  prop.  XXIV,  cor.  1,  that  all 
of  its  sides  are  equal.     A  square  is  thus  seen  to  be  a  special  form  of  a 
rhombus. 


Rhombus. 


Square. 


Prop.  XXIV.]      PARALLELS   AND   PARALLELOGRAMS.  61 

Proposition  XXIV. 

100.  Theorem.  In  any  varallelogram,  (1)  either*  diagonal 
divides  it  into  two  congruent  triangles,  (2)  the  opposite  sides 
are  equal. 


B 

Given  O  ABCD. 

To  prove    that  (1)  A  ABC  s?  A  CD  A, 

(2)         AB  =  DC. 

Proof.    1.  In  the  figure,  Zx  =  Z.x',  Ay  —  Z-y1,  and  AC  =  AC. 

Prop.  (?) 

2.  .-.  A  ABC^A  CD  A,  which  proves  (1).         Prop.  II 

3.  .'.  AB  =  DC,  which  proves  (2).  §  57 
Similarly  for  diagonal  BD,  and  sides  BC  and  AD. 

Corollaries.     1.  If  two  adjacent  sides  of  a  parallelogram 
are  equal,  all  of  its  sides  are  equal. 

For  by  step  3  the  other  sides  are  equal  to  these. 

Hence,  as  stated  in  §  99,  all  of  the  sides  of  a  rhombus  are  equal. 

2.  The  diagonals  of  a  parallelogram  bisect  each  other. 

For  if  diagonal  BD  cuts  AC  at  O,  then,  by  prop.  II,  A  ABO  =  A  CDO, 
whence  AO  =  OC,  and  BO  =  OB. 

In  the  annexed  figure,  if  a  and  a'  are  perpendicular  to  P  and  P' ,  two 
parallels  (prop.  XVII,  cor.  1),  they  are  parallel 
(prop.  XVI,  cor.  3).     Hence  a  =  a',  by  prop.     P 
XXIV.      This  fact    is    usually    expressed   by     p— 
saying, 

3.  Tic o  parallel  lines  are  everywhere  equidistant  from  each 

other. 


62 


PLANE    GEOMETRY 


[Bk. 


Proposition   XXV. 

101.    Theorem.     If  a  convex  quadrilateral  has  tivo  opposite 
sides  equal  and  parallel,  it  is  a  parallelogram. 


Given         a  convex  quadrilateral  ABCD,  with  AB  —  DC,  and 
AB  II  DC. 


To  prove    that  ABCD  is  a  parallelogram. 
Proof.    1.   In  the  figure  Z  x  =  Z  x', 

BD  =  BD,  and  AB  =  DC. 

2.  .\A  ABD  5£  A  GOB,  and  Zy  =  Z  y'. 

3.  .-.  #C  II  JlA 

4.  .'.  ABCD  is  a  O  by  definition. 


Prop.  (?) 

Given 

Prop.  (?) 

Prop.  XVI 


Exercises.  104.  It  is  shown  in  Physics  that  if  two  forces  are  pulling 
from  the  point  J5,  and  the  first  force  is  represented  (see  fig.  to  prop.  XXV) 
by  BA,  and  the  second  by  BC,  the  resultant  (resulting  force)  will  be  rep- 
resented by  the  diagonal  BD.  Show  that,  if  the  two  forces  do  not  pull  in 
the  same  line,  the  resultant  is  always  less  than  the  sum  of  the  two  forces. 

105.  If  two  equal  lines  bisect  each  other  at  right  angles,  what  figure 
is  formed  by  joining  the  ends  ? 

106.  If  the  diagonals  of  a  rectangle  are  perpendicular  to  each  other, 
prove  that  the  rectangle  is  a  square. 

107.  On  the  diagonal  BD  of  O  ABCD,  P  and  Q  are  so  taken  that 
BP  =  QD.  Show  that  APCQ  is  a  parallelogram.  Suppose  P  is  on  DB 
produced,  and  Q  on  BD  produced. 

108.  Prove  that  the  diagonals  of  a  rectangle  are  equal.  Prove  that 
the  diagonals  of  a  rhombus  arc  perpendicular  to  each  other  and  bisect 
the  angles  of  the  rhombus. 


1 


Prop.  XXVI.J     PARALLELS   AND    PARALLELOGRAMS.  63 

Proposition  XXVI. 

102.  Theorem.  If  two  parallelograms  have  two  adjacent 
*ides  and  any  angle  of  the  one  respectively  equal  to  the 
corresponding  parts  of  the  other,   they  are   congruent. 


B'  A  A  B 

Given         LU  ABCD,   A'B'CD',  in   which   AB  =  A'B\  AD  = 
A'D'.  andZD  =  ZD'. 

To  prove    that         O  ABCD  ^  O  A'B'CD'. 

Proof.    1.  ZA  =  ZA'.ZB  =  Z  B\  Z  C  =  Z  C,  for  they  are 
equal  or  supplemental  to  D  or  D'.  Prop.  XXIII 

2.  CD  =  CD',  BC  =  B'C,  for  they  are  equal  to  sides 
that  are  known  to  be  equal.  Pro}).  XXIV 

3.  Apply  lj  ABCD  to  O  A'B'CD'  so  that  AB  coin- 
cides with  its  equal  A'B',  A  falling  on  A'.  Then  AD 
can  be  placed  on  AD'  because  Z  A  =  Z  .!'.  Then 
D  will  fall  on  D',  because  AD  =  A'D'.  Similarly, 
C  will  fall  on  C\  and  C£  on  C'£\ 

Corollaries.  1.  Two  rectangles  are  congruent  if  two  ad- 
jacent sides  of  the  one  are  equal  to  any  two  adjacent  sides  of 
the  other.      (Why  ?) 

2.  Two  squares  are  congruent  if  a  side  of  the  one  equals  >< 
side  of  the  other.     (Why?) 


Exercises.  109.  Is  a  parallelogram  determined  when  any  two  sides 
and  either  diagonal  are  given  ?  when  two  adjacent  sides  and  either  diag- 
onal are  given  ? 

110.    The  angle-  at  either  base  of  an  isosceles  trapezoid  are  equal. 


64  PLANE    GEOMETRY.  [Bk.  I. 

Proposition  XXVII. 

103.  Theorem.  If  there  are  two  pairs  of  lines,  all  of  which 
are  parallel,  and  if  the  segments  cut  off  by  each  pair  on  any 
transversal  are  equal,  then  the  segments  cut  off  on  any  other 
transversal  are  equal  also. 


Given  four  parallels,  of  which  JPX ,  P2  cut  off  a  segment  a, 
and  P3  j  P4  cut  off  an  equal  segment  b,  on  a  trans- 
versal T,  and  cut  off  segments  a',  b',  respectively, 
on  transversal  T. 

To  prove    that  v     a'  =  b'. 

Proof.    1.  Suppose  x  and  y  II  T  as  in  the  figure. 

2.  Then,  in  the  figure,  Z  wx  =  Z  P2T  =  Z  P4T  =  Z  zy. 

Prop.  XVII,  cor.  2 

3.  And    Z  a'w  =  Z  b'z.  Why  ? 

4.  And  x  =  a  =  b  =  y.  Prop.  XXIV 

5.  .'.A  wa'sc  =  A  zb'y,  and  a'  =  b'.    Prop.  XIX,  cor.  7 

Corollaries.  1.  7^  «-  system  of  paral- 
lels cuts  off  equal  segments  on  one  trans- 
versal, it  does  on  every  transversal. 

For  if  a  =  6i  or  b-2 ,  a'  =  b\  or  0%,  respectively, 
and  similarly  for  the  other  transversals. 

2.   The  line  through  the  mid-point  of  one  side  of  a  triangle^ 
parallel  to  another  side,  bisects  the  third  side. 

Draw  a  third  parallel  through  the  vertex.     Then  cor.  1  proves  it. 


Prop.  XXVII.]     PARALLELS   AND   PARALLELOGRAMS.  65 

3.  The  line  joining  the  mid-points  of  two  sides  of  a  triangle 

is  parallel  to  the  third  side. 

For  if  not,  suppose  through  the  mid-point  of  one  of  those  sides  a  line 
is  drawn  parallel  to  the  base  ;  then  this  must  bisect  the  other  side,  by 
cor.  2 ;  .-.it  must  coincide  with  the  line  joining  the  mid-points,  or  else  a 
side  would  be  bisected  at  two  different  points.  (This  is  the  converse  of 
cor.  2.     Draw  the  figure.) 

4.  The  line  joining  the  mid-points  of  two  sides  of  a  triangle 

equals  half  the  third  side.     (Prove  it.) 


Exercises.  111.  The  line  joining  the  mid-points  of  the  non-parallel 
sides  of  a  trapezoid  is  parallel  to  the  bases. 

112.  In  a  right-angled  triangle  the  mid-point  of  the  hypotenuse  is 
equidistant  from  the  three  vertices.     (This  exercise  has 

been  given  before,  and  will  be  repeated,  since  it  is 
important  and  admits  of  divers  proofs.  It  is  here 
easily  proved  by  prop.  XXVII,  cor.  2  ;  for  if  a  =  b, 
then  a'  —  b'  \  but  p  II  e,  .-.  p  ±  a'.  .-.  x  =  b  =  a.)  a'         b' 

113.  The  lines  joining  the  mid-points  of  the  sides  of  a  triangle  divide 
it  into  four  congruent  triangles. 

114.  If  one  of  the  equal  sides  CB  of  an  isosceles  triangle  ABC  is  pro- 
duced through  the  base,  and  if  a  segment  BD  is  laid  off  on  the  produced 
part,  and  an  equal  segment  AE  is  laid  off  on  the  other  equal  side,  then 
the  line  joining  D  and  E  is  bisected  by  the  base.  (Consider  the  cases  in 
which  BD<  CB,  BD  =  CB,  BD>CB.) 

115.  If  the  mid-points  of  the  adjacent  sides  of  any  quadrilateral  are 
joined,  the  figure  thus  formed  is  a  parallelogram.  (Consider  this  theorem 
for  cases  of  concave,  convex,  and  cross  quadrilaterals,  and  for  the  special 
case  of  an  interior  angle  of  180°.) 

116.  The  lines  joining  the  mid-points  of  the  opposite  sides  of  a  quadri- 
lateral bisect  each  other.  Consider  for  the  special  cases  mentioned  in 
ex.  115. 

117.  The  line  joining  the  mid-points  of  the  diagonals  of  a  quadri- 
lateral, and  the  lines  joining  the  mid-points  of  its  opposite  sides,  pass 
through  the  same  point. 

118.  P  and  Q  are  the  mid-points  of  the  sides  AB  and  CD  of  the 
parallelogram  ABCD.  Prove  that  PD  and  BQ  trisect  (divide  into  three 
equal  segments)  the  diagonal  A  C. 


66  PLANE    GEOMETRY.  [Bk.  I. 

EXERCISES. 

119.  "What  is  the  sum  of  the  interior  angles  of  a  polygon  of  20  sides  ? 
of  30  sides  ? 

120.  How  many  degrees  in  each  angle  of  a  regular  polygon  of  12  sides  ? 
of  20  sides  ? 

121.  How  many  sides  has  a  polygon  the  sum  of  whose  interior  angles 
is  48  right  angles  ? 

122.  The  vertical  angle  of  a  certain  isosceles  triangle  is  11°  lo'  20"; 
how  large  are  the  base  angles  ? 

123.  The  exterior  angle  of  a  certain  triangle  is  140°,  and  one  of  the 
interior  non-adjacent  angles  is  a  right  angle ;  how  many  degrees  in  each 
of  the  other  two  interior  angles  ? 

124.  Each  exterior  angle  of  a  certain  regular  polygon  is  10° ;  how 
many  sides  has  the  polygon  ? 

125.  If  P  is  any  point  on  the  side  BC  of  A  ABC,  then  the  greater  of 
the  sides  AB,  AC,  is  greater  than  AP. 

126.  If  the  diagonals  of  a  quadrilateral  bisect  each  other,  prove  that  the 
quadrilateral  is  a  parallelogram.  Of  what  corollary  is  this  the  converse  ? 
Prove  that  the  diagonals  of  an  isosceles  trapezoid  are  equal. 

127.  Conversely,  prove  that  if  the  diagonals  of  a  trapezoid  are  equal, 
the  trapezoid  is  isosceles. 

128.  Is  a  parallelogram  determined  when  its  two  diagonals  are  given  ? 
when  its  two  diagonals  and  their  angle  are  given  ? 

129.  ABC  is  a  triangle  ;  AC  is  bisected  at  M;  BM  is  bisected  at  N; 
AN  meets  BC  at  P;  MQ  is  drawn  parallel  to  AP  to  meet  BC  at  Q. 
Prove  that  BC  is  trisected  (see  ex.  118)  by  P  and  Q. 

130.  A,  C  are  points  on  the  same  side  of  XX' ;  B  is  the  mid-point  of 
AC;  through  A,  B,  C  parallels  are  drawn  cutting  XX  in  A',  B',  C* 
Prove  that  AA'  +  CC  =  2  BB'. 

131.  A  straight  line  drawn  perpendicular  to  the  base  AB  of  an  isos- 
celes triangle  ABC  cuts  the  side  CA  at  B  and  BC  produced  at  E ;  prove 
that  CEB  is  an  isosceles  triangle. 

132.  ABC  is  a  triangle,  and  the  exterior  angles  at  B  and  C  are 
bisected  by  the  straight  lines  BB,  CB  respectively,  meeting  at  B ;  prove 
that  Z.  CBB  +  jZi  =  a  right  angle. 

133.  In  the  triangle  ABC  the  side  BC  is  bisected  at  E,  and  AB  at  G; 
AE  is  produced  to  F  so  that  EF  =  AE,  and  CG  is  produced  to  H  so 
that  Gil  =  CG.     Prove  that  F,  B,  iJare  in  one  straight  line. 


PROBLEMS. 


67 


3.     PROBLEMS. 

104.    Definitions.     A   curve   is   a   line    no    part   of  which   is 
straight. 


Cire«2£/W£% 


105.  A  circle  is  the  finite  portion  of  a  plane  bounded  by  a 
curve,  which   is   called   the   circumference, 
and  is   such   that  all  points  on  that  line 
are  equidistant    from  a  point  within  the 
figure  called  the  center  of  the  circle. 

A  circle  is  evidently  described  by  a  line-seg- 
ment making  a  complete  rotation  in  a  plane, 
about  a  fixed  point  (the  center). 

106.  A  straight  line  terminated  by  the  center  and  the  cir- 
cumference is  called  a  radius,  and  a  straight  line  through  the 
center  terminated  both  ways  by  the  circumference  is  called 
a  diameter  of  the  circle. 


107.  A  part  of  a  circumference  is  called  an  arc. 

Note.  The  above  definitions  are  substantially  those  usually  met  in 
elementary  geometries.  The  student  will  find,  after  leaving  this  subject, 
that  the  word  circle  is  often  used  for  circumference.  Indeed,  there  is 
good  authority  for  so  using  the  word  even  in  elementary  geometry. 

108.  From  the  above  definitions  the  following  corollaries 
may  be  accepted  without  further  proof  : 

1.  A  diameter  of  a  circle  is  equal  to  the  sum  of  two  radii  of 
that  circle. 

2.  Circles  having  the  some  rodii  ore  congruent. 

3.  A  point  is  within  a  circle,  on  its  circumference,  or  outside 
the  circle,  according  as  the  distance  from  that  point  to  the 
center  is  less  than,  equal  to,  or  greater  than,  the  radius. 


68  PLANE   GEOMETRY.  [Bk.  I. 

109.  It  now  becomes  necessary  to  assume  certain  postulates 
relating  to  the  circle. 

Postulates  of  the  Circle. 

1.  All  radii  of  the  same  circle  are  equal,  and  hence  all 
diameters  of  the  same  circle  are  equal. 

2.  If  an  unlimited  straight  line  passes  through  a  point 
within  a  circle,  it  must  cut  the  circumference  at  least  twice, 
and  so  for  any  closed  figure. 

That  it  cannot  cut  the  circumference  more  than  twice  is  proved  in  III, 
prop.  VI,  cor. 

3.  If  one  circumference  intersects  another  once,  it  intersects 
It  again. 

4.  A  circle  has  but  one  center. 

5.  A  circle  may  be  constructed  with  any  center,  and  with  a 
radius  equal  to  any  given  line-segment. 

This  postulate  requires  the  use  of  the  compasses.  As  has  been  stated, 
the  only  instruments  allowed  in  elementary  geometry  are  the  compasses  and 
the  straight-edge,  a  limitation  due  to  Plato.  In  the  more  advanced 
geometry,  where  other  curves  than  the  circle  are  studied,  other  instru- 
ments are  permitted. 

110.  Order  to  be  observed  in  the  solution  of  problems : 
Given.     For  example,  the  angle  A. 

Required.     For  example,  to  bisect  that  angle. 

Construction.  A  statement  of  the  process  of  solving, 
using  only  the  straight-edge  and  compasses  in  drawing  the 
figure  described. 

Proof.  A  proof  that  the  construction  has  fulfilled  the 
requirements.  % 

Discussion.  Any  consideration  of  special  cases,  of  the 
limitations  of  the  problem,  etc.  If  a  problem  has  but  a 
single  solution,  as  that  an  angle  may  be  bisected  but  once, 
the  solution  is  said  to  be  unique. 


Prop.  XXVIII.] 


PROBLEMS. 


69 


Proposition    XXVIII. 
111.    Problem.      To  bisect  a  given  angle. 


Given         the  A  AOB. 

Required    to  bisect  it. 

Construction.    1.  With  center  0  describe  an  arc  cutting  AO&t 
C,  and  OB  at  D.  §  109,  post,  of  O 

2.  Draw  DC.  §  28.  post.  st.  line 

3.  Describe  arcs  with    centers  D,   C,  and  radius  DC. 

Post.  (?) 

4.  Join  their  intersection  P,  with  0.  Post.  (?) 
Then  Z  AOB  is  bisected,  YY'  being  an  axis  of 
symmetry  (§  68). 

Proof.    1.   Draw  DP,  CP '; 

then  OD  =  DC,  DP  =  DC,  DC  =  CP. 

..DP  =  CP. 
OP  =  OP. 
.'.A  OCP^A  ODP, 
and  Z  COP  =  ZPOD. 


§  109,  1 
Ax.  (?) 


But 


COROJ 


Prop.  XII 
An  angle  may  be  divided  into  2,  Jf,  8,  16, 


£",  equal  angles.     (How  ?) 


70  PLANE    GEOMETRY.  [Bk.  I. 

112.  Note  on  Assumed  Constructions.  It  has  been  assumed,  up  to 
prop.  XXVIII,  that  all  constructions  were  made  as  required  for  the 
theorems.  Thus  an  equilateral  triangle  has  been  frequently  mentioned, 
although  the  method  of  constructing  one  has  not  yet  been  indicated  ; 
a  regular  heptagon  has  been  mentioned  in  ex.  93,  and  reference  might 
be  made  to  certain  results  following  from  the  trisection  of  an  angle, 
although  the  solutions  of  the  problems,  to  construct  a  regular  heptagon, 
and  to  trisect  any  angle,  are  impossible  by  elementary  geometry.  But 
the  possibility  of  solving  such  problems  has  nothing  to  do  with  the 
logical  sequence  of  the  theorems  ;  one  may  know  that  each  angle  of  a 
regular  heptagon  is  5  •  ISO0,  whether  the  regular  heptagon  admits  of 
construction  or  not.  Nevertheless,  an  important  part  of  geometry  con- 
cerns itself  with  the  construction  of  certain  figures  —  a  part  of  utmost 
practical  value  and  of  much  interest  to  the  student  of  mathematics. 

113.  Suggestions  on  the  Solution  of  Problems.  The  methods 
of  logically  undertaking  the  solution  of  problems  will  be  dis- 
cussed at  the  close  of  Book  III.  But  at  present  one  method, 
already  suggested  on  p.  35,  should  be  repeated  :  In  attempting 
the  solution  of  a  problem,  assume  that  the  solution  has  been 
accomplished ;  then  analyze  the  figure  and  see  what  results 
follow;  then  reverse  the  process,  making  these  results  precede 
the  solution. 

For  example,  in  prop.  XXVIII,  assume  that  ZAOB  has  been  bisected 
by  YY' ;  if  that  were  done,  and  if  any  point,  P,  on  YY'  were  joined 
to  points  equidistant  from  O,  on  the  arms,  say  C  and  D,  then  A  OCP 
would  be  congruent  to  A  ODP  ;  now  reverse  the  process  and  attempt  to 
make  A  OCP  congruent  to  A  ODP  ;  this  can  be  done  if  OD  can  be  made 
equal  to  OC,  and  PD  to  PC,  because  OP=  OP ;  but  this  can  be  done 
by  §  109,  5. 

This  method  of  attacking  a  problem,  without  which  the 
student  will  grope  in  the  dark,  is  called  Geometrical  Analysis. 


Exercises.  134.  Give  the  solution  of  prop.  XXVIII,  using  P'  instead 
of  P.  Why  is  P  better  than  P'  for  practical  purposes  ?  In  what  case 
would  the  construction  fail  for  the  point  P'  ?  In  that  case  how  many 
degrees  in  Z  A  OB  ? 

135.  In  prop.  XXVIII,  in  what  case  would  P'  fall  below  O  ?  Give 
the  solution  in  that  case,  after  connecting  P'  and  0  and  producing  P'O. 


Prop.  XXIX., 


PROBLEMS. 


71 


Proposition  X  XIX. 

114.    Problem.      To  draw  a  perpendicular  to  a  given  line 
from  a  given  internal  point. 


y 


\ 


o 


\c 


Xp 

Y 

Solution.  This  is  merely  a  special  case  of  prop.  XXVIII, 
the  case  in  which  Z  AOB  is  a  straight  angle.  (Why  ?)  The 
construction  and.  proof  are  identical  with  those  of  prop. 
XXVIII,  and  the  student  should  give  them  to  satisfy  himself 
of  this  fact. 


Exercises.     136.    What  kind  of  a  quadrilateral  is  CPDP'  ?     Prove  it. 

137.  Prove  that  any  point  on  BA  is  equidistant  from  P  and  P'.  Also 
that  any  point  on  YY'  is  equidistant  from  D  and  C. 

138.  In  step  3  of  the  construction  of  prop.  XXVIII  might  the  radius 
equal  two  times  DC  ?  If  so,  complete  the  solution.  Is  there  any  limit 
to  the  length  of  the  radius  in  that  step  ? 

139.  In  the  figure  of  prop.  XXVIII,  suppose  ZPCO  =  130°.  Find 
the  number  of  degrees  in  the  various  other  angles,  not  reflex,  of  the 
figure. 

140.  In  the  figure  of  prop.  XXVIII,  prove  that  the  reflex  angle  BOA 
is  bisected  by  YY',  that  is,  by  PO  produced. 

141.  Also  prove  that  YY'  is  the  perpendicular  bisector  of  DC. 

142.  Also  prove  that  if  0  is  connected  with  P  and  with  P',  OP*  will 
fall  on  OP.     (Prel.  prop.  VIII.) 


72  PLANE    GEOMETRY.  [Bk.  I. 

Proposition  XXX. 

115.    Problem.     To  draw  a  perpendicular  to  a  given  line 
from  a  given  external  point. 


A 


s   /  / 


As — "\ 

Given         the  line  XX'  and  the  external  pt.  P. 
Required    to  draw  a  perpendicular  from  P  to  XX'. 
Construction.    1.  Draw  PR  cutting  XX'.  §  28 

2.  With  center  P  and  radius  PR  const,  a  O. 

§  109,  post,  of  O 

3.  Join  A  and  A',  where  the  circumference  cuts  XX', 
with  P.  §  28,  post,  of  st.  line 

4.  Bisect  Z  A' PA.  Prop.  XXVIII 
The  bisector,  PO,  is  the  required  perpendicular. 

Proof.    1. 


PA  =  PA', 

§  109,  1 

Z  OPA  =  ZA'PO, 

Const.,  4 

and  PO  =  PO. 

2.  .-.AAPO^AA'PO,  Prop.  I 
andZiOP  =  ZPO# 

3.  .-.  Z  ,4 OP  is  a  rt.  Z,  and  PO  _L  XX'.  §§  19,  20 
Note.     The  solution  of  this  problem  is  attributed  to  (Enopides. 


Exercises.  143.  Find  in  a  given  line  a  point  equidistant  from  two 
given  points  A  and  B,  the  mid-point  of  AB  being  also  given. 

144.  Find  a  point  equidistant  from  three  given  points  A,  B,  C,  the 
mid-points  of  A  B  and  BC  being  also  given. 


Prop.  XXXI.] 


PROBLEMS. 


73 


Proposition  XXX I. 
116.    Problem.     To  bisect  a  given  line. 


Given  the  line  AB. 

Required    to  bisect  it. 

Construction.    1.  With,  centers  A.  B,  and  equal  radii  describe 
arcs  intersecting  at  P  and  P'.  Post.  (?) 

2.  Draw  PP'.  Post.  (?) 

3.  Then  PP'  bisects  AB. 


Proof. 


(Let  the  student  give  it.     Draw  AP\  P'B,  BP,  PA.) 


Exercises.  145.  Through  a  given  point  t<:>  draw  a  line  making  equal 
angles  with  the  arms  of  a  given  angle.  Discuss  for  various  relative  posi- 
tions of  the  point. 

146.  To  draw  a  perpendicular  to  a  line  from  one  of  its  extremities, 
when  the  line  cannot  be  produced.     (Ex.  112  suggests  a  plan.) 

147.  Through  two  given  points  on  opposite  sides  of  a  given  line  draw 
two  lines  which  shall  meet  in  the  given  line  and  include  an  angle  which 
is  bisected  by  that  line. 

148.  If  two  isosceles  triangles  have  a  common  base,  the  straight  line 
through  their  vertices  is  a  perpendicular  bisector  of  the  base. 


74  PLANE   GEOMETRY.  [Bk.  I. 

Proposition  XXXII. 

117.    Problem.     From  a  given  point  in  a  given  line  to  draw 
a  line  making  with  the  given  line  a  given  angle. 


C    B 


Given         the  line  AB,  the  point  P  in  it,  and  the  angle  0. 

Required    from  P  to  draw  a  line  making  with  AB  an  angle 
equal  to  Z  0. 

Construction.    1.  On  the  arms  of   ZO  lay  off    OC  —  OB   by 
describing  an  arc  with  center  O  and  any  radius  OC. 

§  109,  post,  of  O 

2.  Draw  CD.  §  28,  post,  of  st.  line 

3.  With  center  P  and  radius  OC,  describe  a  circum- 
ference cutting  PB  in  C.  Post.  (?) 

4.  With  center  C  and  radius  CD,  describe  an  arc  cut- 
ting the  circumference  in  D'.  Post.  (?) 
Draw  PD,  and  this  is  the  required  line. 

Proof.         Draw  CD';  then, 

A  PCD'  and  A  OCD   being   mutually  equilateral, 

Why  ? 
A  PCD  ^  A  OCD, 

and  Z  CPD'  =  Z  COD.  Prop.  XII 


Exercises.     149.    Prove  that,   the   circumferences  must  cut  at  D'   as 
stated  in  step  4. 

150.  See  if  the  solution  of  prop.  XXXII  is  general  enough  to  cover 
the  cases  where  the  Z  O  is  straight,  reflex,  a  perigon. 

151.  From  a  given  point  in  a  given  line  to  draw  a  line  making  an 
angle  supplemental  to  a  given  angle. 


Prop.  XXXIIL]  PROBLEMS.  75 

Proposition  XXXIII. 

118.    Problem.     Through  a  given  point  to  draw  a  line  paral- 
lel to  a  given  line. 


V 




0 


Given         the  line  AB  and  the  point  P. 

Required    through  P  to  draw  a  line  parallel  to  AB. 

Construction.    1.  Join  P  with  any  point.  0,  on  AB. 

§  28.  post,  of  st.  line 

2.  From  P  draw  PC  making  Z  OPC  =  Z  POA.        (?) 
Then  PC  is  the  required  line. 

Proof.  PCWAB.  Why? 

Discussion.    The  solution  fails  if  P  is  on  the  unlimited  line  AB. 


Exercises.  152.  Through  a  given  point  to  draw  a  line  making  a  given 
angle  with  a  given  line.     Notice  that  the  solution  is  not  unique. 

153.  Through  a  given  point  to  draw  a  transversal  of  two  parallels, 
from  which  the  parallels  shall  cut  off  a  given  segment.  Discussion  should 
show  when  there  are  two  solutions,  when  only  one,  when  none. 

154.  To  construct  a  polygon  (say  a  hexagon)  congruent  to  a  given 
polygon. 

155.  Through  two  given  points  to  draw  two  lines  forming  with  a 
given  unlimited  line  an  equilateral  triangle. 

156.  Three  given  lines  meet  in  a  point  ;  draw  a  transversal  such  that 
the  two  segments  of  it,  intercepted  between  the  given  lines,  may  be  equal. 
Is  the  solution  unique  ? 

157.  From  P,  the  intersection  of  the  bisectors  of  two  angles  of  an  equi- 
lateral triangle,  draw  parallels  to  two  sides  of  the  triangle,  and  show  that 
these  parallels  trisect  (see  ex.  118)  the  third  side. 


fl  0 


76  PLANE    GEOMETRY.  [Bk.  I. 

Proposition  XXXIV. 
119.    Problem.     To    construct  a  triangle,   given    the    three 


,. 


Given         a,  b,  c,  three  sides  of  a  triangle. 

Required    to  construct  the  triangle. 

Construction.    1.  With  the  ends  of  b  as  centers,  and  with  radii 
a,  c,  describe  circumferences.  §  109 

2.  Connect  either  point  of  intersection  of  these  circum- 
ferences with  the  ends  of  b.  §  28 
Then  is  the  required  A  constructed. 

Proof.         It  was  constructed  on  b,  and  the  other  sides  equal  a,  c. 

§  109,  1 

Discussion.    If    the    two    circumferences    do  not   intersect,    a 

solution   is    impossible,  for   then   either   a  >  b  +  c, 

a  =  b  +  c,  a  =  c  —  b,  or  a  <  c  —  b,  and  in  none  of 

.  these  cases  is  a  triangle  possible. 

Prop.  VIII  and  cor. 
Corollary.      To  construct  an  equilateral  triangle  on  a  given 
line-segment. 

The  first  proposition  of  Euclid's  "Elements  of  Geometry."  Euclid 
proceeded  upon  the  principle  of  logical  sequence  of  propositions,  with 
no  attempt  at  grouping  the  theorems  and  the  problems  separately.  He 
found  this  corollary  (a  problem)  the  best  proposition  with  which  to  begin 
his  system. 

Exercise.  158.  In  a  given  triangle  inscribe  a  rhombus,  having  one  of 
its  angles  coincident  with  a  given  angle  of  the  triangle,  and  the  other 
three  vertices  on  the  three  sides  of  the  triangle. 


Prop.  XXXV.]  PROBLEMS.  77 

Propositiox  XXXV. 

120.    Problem.     To  construct  a  triangle,  given  two  sides  and 
the  included  angle. 


b 

Given         the  sides  a,  b,  and  the  included  angle  k. 

Required    to  construct  the  triangle. 

Construction.    1.  From  either  end  of  b  draw  a  line  making  with 
b  the  angle  k.  Prop.  XXXII 

2.  On  that  line  mark  off  a  by  describing  an  arc  of 
radius  a.  §  109 

3.  Join  the  point  thus  determined  with  the  other  end 
of  b.  §  28 
Then  the  triangle  is  constructed. 

Proof.         By  step  2  the  line  marked  off  equals  a,  and  by  step  1 
Z-b  =  Z.  k,  and  it  is  constructed  on  b. 


Exercises.  159.  To  trisect  a  right  angle.  (Construct  an  equilateral 
triangle  on  one  arm.) 

160.  On  the  side  AC  of  A  ABC  to  find  the  point  P  such  that  the  par- 
allel to  AB,  from  P,  meeting  BC  at  Z>,  shall  have  PD  =  A  P. 

161.  To  construct  a  triangle,  having  given  two  angles  and  the  perpen- 
dicular from  the  vertex  of  the  third  angle  to  the  opposite  side. 

162.  Draw  a  line  parallel  to  a  given  line,  so  that  the  segment  inter- 
cepted between  two  other  given  lines  may  equal  a  given  segment. 

163.  Given  the  three  mid-points  of  the  sides  of  a  triangle,  to  construct 
the  triangle. 

164.  Through  a  given  point  P  in  an  angle  AOB  to  draw  a  line,  termi- 
nated by  OA  and  07?,  and  bisected  at  P.  (Through  P  draw  a  II  to  BO 
cutting  OA  in  X ;  on  XA  lay  off  XY  =  OX ;  draw  YP.) 


'0/li-o* 


78  PLANE   GEOMETRY.  [Bk.  I. 

Proposition  XXXVI. 

121.    Problem.     To  construct  a  triangle,  given  two  sides  and 
the  angle  opposite  one  of  them. 


Given  two  sides  of  a  triangle,  a,  b,  and  Z  k  opposite  a. 

Required    to  construct  the  triangle. 

Construction.  1.  At  either  end  of  b  draw  a  line  making  with  b 
an  angle  equal  to  Z  k.  Prop.  (?) 

2.  With  the  other  end  of  b  as  a  center,  and  a  radius  a, 
describe  a  circumference.  Post.  (?) 

3.  Join  the  points  where  the  circumference  cuts  the 
line  of  step  1,  with  the  center.  Post.  (?) 
Then  the  triangle  is  constructed. 

Proof.  For  it  has  the  given  side  b,  and  the  given  Z  k,  and 
the  lines  of  step  3  equal  a.  §  109,  1 

Discussion.  If  the  circumference  cuts  the  line  twice,  two  solu- 
tions are  possible,  and  the  triangle  is  ambiguous 
(see  prop.  XIV).  If  it  touches  the  line  without 
cutting  it,  what  about  the  solution  ?  If  it  does  not 
meet  the  line,  no  solution  is  possible.  If  Z  k  is 
right  or  obtuse,  or  if  a<£b,  only  one  solution  is 
possible  (prop.  XIX,  cor.  5), 

Draw  a  figure  for  each  of  these  cases,  and  show  from  the  drawings 
that  the  statements  made  in  the  discussion  are  true. 


Exercise.  165.  XX%  YY',  are  two  given  lines  through  0,  and  P  is  a 
given  point ;  through  P  to  draw  a  line  to  XX',  which  shall  he  hisected 
by  YY'.  Investigate  for  various  positions  of  P.  as  where  P  is  within 
the  ZXOY.  the  Z  YOX\  on  OY,  or  on  OX. 


Prop.  XXXVII.]  PROBLEMS.  79 

Proposition  XXXVII. 

122.    Problem.      To  construct  a  triangle,  given  two  angles 
and  the  included  side. 


X 


A  A  B  B  A  B 

Given         two  angles,  A,  B,  and  the  included  side  AB. 

Required    to  construct  the  triangle. 

Construction.    1.  From  A  draw  A X  making  with  AB  an  angle 
equal  to  Z  A.  Prop.  XXXI 1 

2.   Similarly,  from  B  draw  BY,  making  an  angle  equal 
to  Z  B."  Prop.  XXXII 

C  being  the  intersection  of  AX,  BY,  then  ABC  is 
the  required  A. 

Proof.         (Let  the  student  give  it.) 

Discussion.    If  AX,  BY,  do  not  intersect,  what  follows  ? 


Proposition  XXXVIII. 

123.  Problem.  To  construct  a  triangle,  given  two  angles 
and  a  side  opposite  one  of  them. 

Solution.  Subtract  the  sum  of  the  angles  (found  by  prop. 
XXXII)  from  180°  and  thus  find  the  third  angle  (prop.  XIX). 
The  problem  then  reduces  to  prop.  XXXVII. 


Proposition  XXXIX. 

124.    Problem.      To  construct  a  square  on  a  given   line  as 
a  side. 


80  PLANE    GEOMETRY.  [Bk.  I. 


4.    LOCI   OF   POINTS. 

125.  The  place  of  all  points  satisfying  a  given  condition  is 
called  the  locus  of  points  satisfying  that  condition. 

Indeed,  the  word  locus  (Latin)  means  simply  place  (English,  locality, 
locate,  etc.)  ;  the  plural  is  loci. 

For  example,  if  points  are  on  this  page  and  are  one  inch  from  the  left 
edge,  their  locus  is  evidently  a  straight  line  parallel  to  the  edge. 

Furthermore,  the  locus  of  points  at  a  given  distance  r  from  a  fixed 
point  0  is  the  circumference  described  about  0  with  a  radius  r.  This 
statement,  although  very  evident,  is  made  a  theorem  (prop.  XL)  because 
of  the  frequent  reference  to  it. 

Of  course  in  this  discussion,  as  elsewhere  in  Books  I— V, 
the  points  are  all  supposed  to  be  confined  to  one  plane. 

In  Plane  Geometry  the  loci  considered  will  be  found  to  con- 
sist of  one  or  more  straight  or  curved  lines. 

It  is  a  common  mistake  to  assume  that  a  locus,  which  one  is  trying  to 
discover,  consists  of  a  single  line.  It  may  consist  of  two  lines,  as  in  prop. 
XLII. 

126.  In  proving  a  theorem  concerning  the  locus  of  points  it 
is  necessary  and  sufficient  to  prove  two  things  : 

1.  That  any  point  on  the  supposed  locus  satisfies  the  condition; 

2.  That  any  point  not  on  the  supposed  locus  does  not  satisfy 
the  condition. 

For  if  only  the  first  were  proved,  there  might  be  some  other 
line  in  the  locus  ;  and  if  only  the  second  were  proved,  the  sup- 
posed locus  might  not  be  the  correct  one. 


Exercise.  166.  State,  without  proof,  what  is  (1)  the  locus  of  points 
\  in.  from  a  given  straight  line  ;  (2)  the  locus  of  points  equidistant  from 
two  parallel  lines. 


k 


Prop.  XL.]  LOCI    OF   POINTS.  81 

Proposition  XL. 

127.  Theorem.  The  locus  of  points  at  a  given  distance 
from  a  given  point  is  the  circumference  described  about  that 
point  as  a  center,  with  a  radius  equal  to  the  given  distance. 


Given         the  point   0,  the  line  r.  and  the  circumference   C 
described  about    0  with  radius  /■. 

To  prove    that  C  is  the  locus  of  points  r  distant  from  0. 

Proof.    1.   Let  Px,  P2,  P3  be  points  on  C\  within  the  circle,  and 
without  the  circle,  respectively. 
Let  OP2  produced  meet  C  in  B.  and  OP3meet  Cm  A. 

2.  Then  OP1  =  OB  =  OA  =  r,  §  109,  1 
and  OP,  <  OB,  and  OPz  >  OA.  Ax.  8 

3.  .*.  any  point  on  C  is  r  distant  from  0,  and  any  point 
not  on  C  is  not  r  distant  from  0. 


Exercises.  167.  Has  it  been  proved  in  prop.  XL  that  the  required 
locus  may  not  be  merely  the  arc  cut  off  by  r  and  OP\  ?     If  so,  where  ? 

168.  What  is  the  locus  of  points  at  a  distance  of  i  in.  from  the  above 
circumference,  the  distance  being  measured  on  a  line  through  0? 

169.  Lighthouses  on  two  islands  are  10  miles  apart ;  show  that  there 
are  two  points  at  sea  which  are  exactly  12  miles  from  each. 

170.  How  would  you  find,  by  the  intersection  of  two  loci,  a  point  on 
this  page  1  in.  from  0  in  the  above  figure,  and  3  in.  from  the  right  edge 
of  the  paper  ? 


82 


PLANE    GEOMETRY. 


[Bk.  I. 


Proposition  XLI. 

128.  Theorem.  The  locus  of  points  equidistant  from  two 
given  points  is  the  perpendicular  bisector  of  the  line  joining 
them. 

Y 


Given         two  points  Xand  X\  and  IT'  _L  XX'  at  the  mid- 
point 0. 

To  prove    that  YY'  is  the  locus  of  points  equidistant  from  X 
and  X\ 

Proof.    1.  Let  P  be  any  point  on  YY',  and  P'  be  any  point  not 
on  YY'. 
Draw  PX,  PX',  P'X,  P'X'. 

2.  Then  PX=  PX',  Prop.  XX,  cor.  5 
and  P'X'  >  P'X.  Prop.  XX,  cor.  2 

3.  Hence  any  point  on  YY'  is  equidistant  from  X  and 
X',  but  any  point  not  on  YY'  is  unequally  distant 
from  X  and  A"'. 

.-.  YY'  is  the  locus.  §  125,  def. 


Exercises.  171.  Required  to  find  a  point  which  is  1  in.  from  X  and 
|  in.  from  X'  in  the  above  figure.     Is  there  more  than  one  such  point  ? 

172.  Required  to  find  a  point  which  is  equidistant  from  X  and  X'  in 
the  above  figure,  and  1  in.  from  0.     Is  there  more  than  one  such  point  ? 


Prop  XLII. 


LOCI    OF   POINTS. 


83 


Proposition  XLII. 

129.    Theorem.      The  locus  of  points  equidistant  from  two 
given  lines  consists  of  the- bisectors  of  their  included  angles. 


Given  OA  and  OB,  two  lines  intersecting  at  0,  and  XX' 

and  YY'  the  bisectors  of  the  angles  at  0. 

To  prove  that  XX'  and  YY'  form  the  locus  of  points  equidis- 
tant from  OA  and  OB. 

Proof.  1.  Let  Q  be  any  point  on  neither  XX'  nor  YY';  let 
QB  _L  OB,  QA  _L  OA,  QA  cut  OX  in  P,  PA'  _L  0#. 
Draw  QA'. 

Since  $  may  be  moved,  P  may  be  considered  as  any 
point  on  OX. 

2.  Then  A  OAP  ^  A  OA'P, 

and  JP  =  A' P.  Prop.  XIX,  cor.  7 

3.  Also,        A'P  +  PQ>  A'Q  >  BQ.  Why  ? 

4.  .-.  AQ,ovAP  +  PQ>  BQ.  Why? 

5.  .\  any  point  P  on  XX'  (or  on  77')  is  equidistant 
from  OA  and  0^,  but  any  point  Q  on  neither  XXi 
nor  77'  is  unequally  distant  from  OA  and  OB. 


84  PLANE    GEOMETRY.  [Bk.  I. 

Corollaries.  1.  If  the  given  lines  are  parallel,  the  locus 
is  a  parallel  midway  between  them.     (Prove  it.) 

The  student  should  imagine  the  effect  of  keeping  points  A,  A'  fixed, 
and  moving  0  farther  to  the  left.  YY'  moves  with  0,  but  XX'  keeps 
its  position  as  the  lines  approach  the  condition  of  being  parallel. 

2.  The  locus  of  points  at  a  given  distance  from  a  given  line 
consists  of  a  pair  of  parallels  at  that  distance,  one  on  each  side- 
of  the  fixed  line.      (Prove  it.) 

130.    Definitions. 

Three  or  more  lines  which  Three  or  more  points  which 

meet  in  a  point  are  said  to  be  lie  in   a  line  are  said  to  be 

concurrent.  collinear. 


Proposition  XLIII. 

131.    Theorem.     The  perpendicular    bisectors   of  the  three 
sides  of  a  triangle  are  concurrent. 


Given         a  triangle  of  sides  a,  b,  c,  and  x,  y,  z  their  respective 
perpendicular  bisectors. 

To  prove    that  x,  y,  z  are  concurrent. 

Proof.    1.  x  and  y  must  meet  as  at  P.  Prop.  XVII,  cor.  4 

2.  Then  P  is  equidistant  from  B  and  C,  and  C  and  A. 

Prop.  XLI 

3.  .'.  P  is  on  the  perpendicular  bisector  of  c  ;       Why  ? 
i.e.  z  passes  through  P. 


Prop.  XLIIL]  LOCI    OF   POINTS.  85 

Corollaries.  1.  The  point  equidistant  from  three  non- 
coUinear  points  is  the  intersection  of  the  perpendicular  bisectors 
of  any  two  of  the  lines  joining  them. 

Step  2. 

2.  There  Is  one  circle,  and  only  one,  whose  circumference 
passes  through  three  non-collinear  points. 

Let  A,  B,  C  be  the  three  points.  Then  by  step  2  they  are  equidistant 
from  P,  the  intersection  of  x  and  y. 

And  v  x  and  y  contain  all  points  equidistant  from  A,  B,  and  C,  and 
can  intersect  but  once,  there  is  only  one  point  P. 

And  v  there  is  only  one  center  and  one  radius,  there  is  one  and  only 
one  circle. 

3.  Circumferences  having  three  points  in  common  are  iden- 
tical. 

Otherwise  cor.  2  would  be  violated. 

4.  If  from  a  point  more  than  two  lines  to  a  circumference 
are  equal,  that  point  Is  the  center  of  the  circle. 

For  suppose  a  circumference  through  A,  B.  C,  and  suppose  PA  =  PB 
=  PC. 

Now  with  center  P  and  radius  PA  a  circumference  can  be  described 
through  A,  B,  C,  because  it  is  given  that 

PA=PB  =  PC.  §  108,  cor.  3 

And  this  is  identical  with  the  given  circumference. 

Prop.  XLIIL  cor.  3 

.-.  its  center  must  be  identical  with  the  given  center,  since  a  O  cannot 

have  two  centers.  §  109,  4 

Exercises.  173.  The  proof  of  prop.  XLIII  is,  of  course,  the  same  if 
the  triangle  is  right-angled  or  obtuse-angled.  The  figures,  however, 
show  interesting  positions  for  P ;  consider  them. 

174.  Required  to  find  a  point  at  a  given  distance  d  from  a  fixed  point 
0,  and  equidistant  from  two  given  intersecting  lines.  How  many  such 
points  can  be  found  in  general  ? 

175.  Required  to  find  a  point  equidistant  from  two  given  intersecting 
lines,  and  equidistant  from  two  given  points.  How  many  such  points 
can  be  found  in  general  ? 


86 


PLANE    GEOMETRY. 


[Bk.  I. 


Proposition  XL IV. 

132.    Theorem.     The  bisectors  of  the  interior  and  exterior 
angles  of  a  triangle  are   concurrent  four    times  by  threes. 


.-'/^ 


F 

>-'    A. 


XR 


Given         the  A  ABC,  and  the  bisectors  of  the  interior  and 
exterior  angles,  lettered  as  in  the  figure. 

To  prove    that  these  bisectors  are  concurrent  four   times   by 
threes ;  that  is,  3  meet  at  P1?  3  at  P2,  etc. 

Proof.    1.  v  Z  CAM  >  Z  CBM, .'.  Z  GAM  >  Z  HBM.  Prop.  V 

2.  .-.  AG  and  BR  meet  as  at  P3.       Prop.  XVII,  cor.  3 

3.  Z  HBM+  Z  P^  <  Z  5  +  Z  A  <  180°.  Prop.  XIX 
.'.  BH  and  .^P  meet  as  at  Pv       Prop.  XVII,  cor.  3 

4.  PP_L  PiT,  and  AG  _L  JP,  Prel.  prop.  IX 
.*.  PP  and  AG  meet  as  at  P4.        Prop.  XVII,  cor.  4 

5.  Also,  Px  is  equidistant  from  a  and  c,  from  c  and  b, 
and  .'.  from  a  and  &,  Prop.  XLII 
,\PX  lies  on  CT.     Similarly  for  P4.        Prop.  XLII 

6.  Similarly,  P2  and  P3  lie  on  CN.  .'.  the  four  points 
Px,  P2,  P3,  P4,  are  points  of  concurrence  of  the 
bisectors. 


Prop.  XLV.]  LOCI    OF  POINTS.  87 

Proposition  XLV. 

133.    Theorem.      The  perpendiculars  from  the  vertices  of  a 
triangle  to  the  opposite  sides  are  concurrent. 


A\ Z 


X' 


Given  the  A  ABC. 


To  prove    that   the    perpendiculars    from    A,  B,  C,  to  a,  b,  c, 
respectively,  are  concurrent. 

Proof.    1.  Through  A,  B,  C,  respectively,  suppose  B'C  II  CB, 
A'C  II  CA,  A'B'  II  BA. 

2.  Then  ABCB'  and  ABAC  are  UJ.  Def.  O 

3.  .-.  B'C  =  AB=  CA'.  and  C  is  the  mid-point  of  B'A\ 

Prop.  XXIV;  ax.  1 

4.  Similarly,  A  and  B  are  mid-points  of  B'C\  CA'. 

5.  If  AX,  BY,  CZ±B'C,  CA',  A'B',  respectively,  they 
are  concurrent,  as  at  0.  Prop.  XLIII 

6.  And  they  are  also  the  perpendiculars  from  A,  B,  C 
to  a,  b,  c.  Prop.  XVII,  cor.  1 

Note.     The  theorem  is  due  to  Archimedes. 

134.    Definition.     To  trisect  a  magnitude    is    to   cut  it  into 
three  equal  parts. 

Exercise.     176.    In  prop.  XLIV  suppose  C  moves  down  to  the  side  c. 
What  becomes  of  Pi,  P2,  P3,  P4  ? 


88  PLANE   GEOMETRY  [Bk.  I. 

Proposition  XLVI. 

135.    Theorem.     The  medians  of  a  triangle  are  concurrent 
in  a  trisection  point  of  each. 


Given         the  A  ABC  and  the  medians  BY,  AX,  intersecting 
at  0. 

To  prove    that  (1)  the  median  from  C  must  pass  through  0, 
(2)   OX  =  I  AX,   OY=iBY,  etc 

Proof.    1.  Suppose     CO    drawn,    and    produced    indefinitely, 
cutting  AB  at  Z. 

2.  Suppose  AP  II  OB ;   CO  must  cut  AP,  as  at  P.    §  85 

3.  Draw  PB.     Then  V  CY=  YA,  .'.  CO  =  OP. 

Prop.  XXVII,  cor.  2 

4.  And  V  CO  =  OP,  and  CX=  XB,  .'.  OX  II  PB. 

Prop.  XXVII,  cor.  3 

5.  .'.  APBO  is  &EJ,  AZ  =  ZB.  and  OZ  =  ZP. 

Prop.  XXIV,  cor.  2 

6.  . ' .  OZ  is  a  median,  and  it  passes  through  0. 

7.  And  v  0Z=  £  OP,  .'.  O^  =  £  CO,  or  J  CZ.     Simi- 
larly for  OF  and  OA: 


Exercise.     177.    The  sum  of  the  three  medians  of  a  triangle  is  greater 
than  three-fourths  of  its  perimeter. 


Sacs.  13&-139.]  LOCI   OF  POINTS.  89 

136.  Definitions.  The  point  of  concurrence  of  the  perpen- 
dicular bisectors  of  the  sides  of  a  triangle  is  called  the  cir- 
cumcenter  of  the  triangle.      (Prop.  XLIII.) 

The  reason  will  appear  later  when  it  is  shown  that  this  point  is  the 
center  of  the  circum-scvihed  circle.     (See  Table  of  Etymologies.) 

137.  The  point  of  concurrence  of  the  bisectors  of  the  interior 
angles  of  a  triangle  is  called  the  in-center  of  the  triangle ;  the 
points  of  concurrence  of  the  bisectors  of  two  exterior  angles 
and  one  interior  are  called  the  ex-centers  of  the  triangle. 
(Prop.  XLIV.) 

It  will  presently  be  proved  that  the  in-center  is  the  center  of  a  circle, 
in-side  the  triangle,  just  touching  the  sides ;  and  that  the  ecc-centers  are 
centers  of  circles,  out-side  the  triangle,  just  touching  the  three  lines  of 
which  the  sides  of  the  triangle  are  segments.  Hence  the  names  in-center 
and  ex-center. 

138.  The  point  of  concurrence  of  the  three  perpendiculars 
from  the  vertices  to  the  opposite  sides  is  called  the  orthocenter 
of  the  triangle.     (Prop.  XLY.) 

139.  The  point  of  concurrence  of  the  three  medians  of  a  tri- 
angle is  called  the  centroid  of  that  triangle.     (Prop.  XLYI.) 

It  is  shown  in  Physics  that  this  point  is  also  the  center  of  mass,  or 
center  of  gravity  of  the  plane  surface  of  the  triangle.  It  is,  therefore, 
sometimes  called  bv  those  names. 


Exercises.     178.    If  a   triangle  is  acute-angled,  prove  that  both  the 
circumcenter  and  the  orthocenter  lie  within  the  triangle. 

179.  In  prop.  XLVI,  if  X,  Y,  Z  be  joined,  prove  that  the  A  XYZ 
will  be  equiangular  with  the  A  ABC. 

180.  Is  there  any  kind  of  a  triangle  in  which  the  in-center,  circum- 
center. orthocenter,  and  centroid  coincide  ?     If  so,  what  is  it  ?     Prove  it. 

181.  In  the  figure  of  prop.  XLVI,  connect  X,  F,  Z,  and  prove  that  0 
is  also  the  centroid  of  A  XYZ. 

182.  In  ex.  179,  prove  that  if  the  mid-points  of  the  sides  of  A  XYZ 
are  joined,  0  is  also  the  centroid  of  that  triangle  ;  and  so  on*. 


BOOK   II.  — EQUALITY   OF   POLYGONS. 


1.    THEOREMS. 

140.  Definitions.     Two  polygons  are  said  to  be  adjacent  if 
they  have  a  segment  of  their  perimeters  in  common. 

141.  Suppressing  the  common  segment  of  the  perimeters  of 
two  adjacent  polygons,  a  polygon  results  which 
is  called  the  sum  of  the  two  polygons.     Simi- 
larly for  the  sum  of  several  polygons,  and  for 
the  difference  of  two  overlapping  polygons. 

142.  Surfaces  which  may  be  divided  into  the 


same  number  of  parts  respectively  congruent,  or  which  are  the 
differences  between  congruent  surfaces,  are  said  to  be  equal. 

This  property  is  often  designated  by  the  expressions  equivalent,  equal 
in  area,  of  equal  content,  etc.;  but  the  use  of  the  word  congruent,  for 
identically  equal,  renders  the  word  equal  sufficient. 

The  definition  is  more  broadly  treated  in  Book  V. 


143.  The   altitude  of   a  trapezoid  is 

the   perpendicular    distance    between 
the  base  lines. 

Hence  a  trapezoid  can  have  but  one  alti- 
tude, a,  unless  it  becomes  a  parallelogram. 

144.  The  altitude  of  a  triangle  with 
reference  to  a  given  side  as  the  base, 
is  the  distance  from  the  opposite  ver- 
tex to  the  base  line. 

Hence  a  triangle  can  have  three  distinct 
altitudes,  viz.  au  a2,  a3,  in  the  figure. 

90 


Prop.  I.j  EQUALITY    OF   POLYGONS.  91 

Proposition  I. 

145.    Theorem.     Parallelograms   on   the    same  base   or   on 
equal  bases  and  between  the  same  parallels  are  equal. 


Given  UJ  ABCD,    ABC'D',    on   the    same    base   AB,    and 

between  the  same  parallels  P.  P'. 

To  prove  that  O  ABCD  =  O  A  BCD'. 

Proof.    1.  AD  =  BC,  AD'  =  BC.  DC  =  AB  =  D'C.      Why  ? 

2.  In  Fig.  1,  adding  CD'.  DD'  =  CC.  Ax.  2 

3.  .'.ABC'C^AAD'D.  Why? 

4.  But  ABC'D  =  ABC'D. 

. ' .  O  AB  CD  =  O  AB  C  !D.  Ax.  3 

Similarly  for  Figs.  2  and  o.  In  Fig.  2,  CZ)'  has  become 
zero ;  in  Fig.  3,  it  has  become  negative. 

The  meaning  of  ••  between  the  same  parallels  "  is  apparent. 

Corollaries.     1.    A  parallelogram  equals  a  rectangle  of  the 

same  base  and  the  same  altitude.     (Why  ?) 

2.  Parallelograms  hawing  equal  bases  and  equal  altitudes 
are  equal.      (Why  ?) 

3.  Of  two  parallelograms  having  equal  altitudes,  that  is  the 
greater  ivhich  has  the  greater  base ;  and  of  two  having  equal 
bases,  that  is  the  greater  which  has  the  greater  altitude.    (Why  ?) 

4.  Equal  parallelograms  on  the  same  base  or  on  equal  bases 
have  equal  altitudes. 

Law  of  Converse,  §  73,  after  cors.  2  and  3.     Give  it  in  full. 


92  PLANE    GEOMETRY. 


Proposition  II. 


[Be.  il 


146.    Theorem.      Triangles  on  the  same  base    or  on  equal 
bases  and  between  the  same  parallels  are  equal. 


Given  A  ABC,  ABC  on  the  base  AB,  and   between  the 

same  parallels  AB,   C'C. 

To  prove    that  A  ABC  =  A  ABC. 

Proof.    1.   In  the  figure,  suppose  AD  II  BC,  BD'  II  AC. 

Then  ABCD,  ABD'C  are  equal  UJ.  Why  ? 

2.  And,  since  A  ABC,  ABC  are  their  halves, 

I,  prop.  XXIV 
.\AABC=AABC.  Ax.  7 

Corollaries.  1.  A  triangle  equals  half  of  a  parallelogram, 
or  half  of  a  rectangle,  of  the  same  base  and  the  same  altitude 
as  the  triangle. 

By  step  2,  and  prop.  I,  cor.  1. 

2.  Triangles  having  equal  bases  and  equal  altitudes  are  eq ual. 

3.  Of  two  triangles  having  equal  altitudes,  that  is  the  greater 
which  has  the  greater  base  ;  and  of  two  having  equal  bases,  that 
is  the  g?*eater  ivhich  has  the  greater  altitude.      (Why  ?) 

4.  Equal  triangles  on  the  same  base  or  on  equal  bases  have 
equal  altitudes.      (Why  ?) 

Note.  In  props.  I  and  II  if  the  figures  are  on  equal  bases  they  can 
evidently  be  placed  on  the  same  base.  Hence  the  proofs  given  are 
sufficient. 


Prop.  III.]  EQUALITY    OF  POLYGOXS.  93 

Proposition  III. 

147.  Theorem.  A  trapezoid  is  equal  to  half  of  the  rect- 
angle whose  base  is  the  sum  of  the  two  parallel  sides,  and 
whose  altitude  is  the  altitude  of  the  trapezoid. 


b\i L 

B  D 

Given  the  trapezoid  ABCD. 

To  prove    that  ABCD  equals  half  of  a  rectangle  with  the  same 
altitude,  and  with  base  equal  to  AB  -f-  DC. 

Proof.    1 .  About  0,  the  mid-point  of  B  C.  revolve  AB  CD  through 
180°  to  the  position  A'CBD',  leaving  its  original  trace. 

2.  Then,  '■'  Zc'  =  Z c,  and  Zb  +  Zc  =  st.  Z, 
.-.  Z  b  +  Z  c'  =  Z  b  +  Z  c  =  st.  Z, 

and  .  • .  ABD'  is  a  st.  line.  §  14,  clef.  st.  Z 

.  Similarly,  DC  A'  is  a  st.  line. 

3.  Also,    v  ZD'  =  ZD,  and  ZA  +  ZD=st.  Z. 
.'.ZA  +  ZD'  =  ZA  +  ZD  =  st.  Z, 

.'.  D'A'  li  AD,  I,  prop.  XVI,  cor.  2 

and  .-.  ADA'D  is  a  O.  §  97,  def.  O 

4.  The  base  of  the  O  =  AB  +  DC, 

and  the  O  =  2  •  ^CD.  Why  ? 

5.  .'.  ^LBOZ>  =  £  O  =  £  required  □.     Prop.  I,  cor.  1 


Exercises.    183.    P  is  any  point  within  O  ABCD.    Prove  that  A  PAB 
+  A  PCD  =  £  O  ABCD.     Suppose  P  is  outside  of  LJ  ABCD. 

184.    A  quadrilateral  equals  a  triangle  of  which  two  sides  equal  the    a  \ 
diagonals  of   the  quadrilateral,  and   the  included  angle  of   those   sides    <f 
equals  the  included  angle  of  the  diagonals. 


94 


PLANE   GEOMETRY. 


[Bk.  II. 


Proposition  IV. 


148.  Theorem.  If  through  a  point  on  a  diagonal  of  a  par- 
allelogram parallels  to  the  sides  are  drawn,  the  parallelo- 
grams on  opposite  sides  of  that  diagonal  are  equal. 


Given 


A  E  B 

O  AB  CD,  and  through  P,  a  point  on  A  C,  the  lines 
GF  II  AB,  TIE  II  DA,  and  the  parts  lettered  as  in  the 
figure. 

To  prove    that  b  =  b'. 

Proof.    1.  a  +  b  -f-  c  =  a'  +  V  +  c', 


,\b=bl 


I,  prop.  XXIV 
Ax.  3 


149.  Definitions.  Since  all  rectangles  which  have  two  adja- 
cent sides  equal  to  two  given  lines  a,  b,  are  congruent  (I,  prop. 
XXVI,  cor.  1),  any  such  rectangle  is  spoken  of  as  the  rectangle 
of  a  and  b. 

This  is  indicated  by  the  symbol  ab,  or,  if  the  adjacent  sides  are  AB 
and  CD,  by  the  symbol  AB  ■  CD.  These  symbols  are  read  "The  rect- 
angle of  a  and  6,"  "The  rectangle  of  AB  and  CD,1'  or,  briefly,  "The 
rectangle  <z&,"  "The  rectangle  AB  (pause)  CD."  Since  there  is  no  mul- 
tiplication of  lines  by  lines,  by  any  definition  thus  far  known  to  the  stu- 
dent, the  readings  "a  times  &,"  "  AB  times  CD"  are  not  recommended. 

In  like  manner,  any  square  whose  side  is  equal  to  a  given 
line  is  spoken  of  as  the  square  on  (or  of)  that  line. 

The  square  on  a  line  AB  is  indicated  by  the  symbol  AH1  ;  on  a  line  a 
by  the  symbol  a2  ;  read  "The  square  on  (or  of)  AB,"  or,  briefly,  "iB- 
square  "  ;  and  similarly  for  a. 

Squares,  rectangles,  and  polygons  in  general  are  often  designated  by 
the  letters  of  two  vertices  not  consecutive. 


Sec.  150.]  EQUALITY   OF  POLYGONS.  95 

150.  A  point  in  a  line-segment  is  said  to  divide  it  internally ; 
a  point  in  a  produced  part  of  a  line-segment  is  said  to  divide 
it  externally. 

In   the    figure,  AB  is  di- ^ - ^ 

vided    internally  at  P,  and 

externally  at  P'.     AP,  PB  are  called  segments  of  AB:  and 

AP'.  P'B  are  also  called  segments  of  AB. 

The  propriety  of  calling  AP,  PB,  and  AP",  P'B,  segments  of  AB  is 
apparent,  since  AP  +  PB  =  AB,  and  also  AP"  +  P'B  (which  is  negative) 
=  AB.  

Exercises.  185.  If  the  sides  BC,  CA,  AB,  of  A  ABC,  are  produced 
to  X,  Y,  Z,  respectively,  so  that  CX  =  BC,  AY  =  CA,  BZ  =  AB,  prove 
that  A  XYZ  =  7  •  A  ABC. 

186.  The  medians  of  a  triangle  divide  it  into  six  equal  triangles.  (In 
what  kind  of  a  triangle  are  the  six  triangles  congruent  ?) 

187.  Prove  prop.  Ill  by  bisecting  BC  at  0,  drawing  DO  to  meet  AB 
produced  at  D',  and  proving  that  A  BIYO  ^  A  CDO,  that  A  AB 'D  = 
trapezoid,  etc. 

188.  Discuss  prop.  IV  when  P  moves  to  C  ;  through  C  on  AC  pro- 
duced. 

189.  If  two  equal  triangles  are  on  opposite  sides  of  a  common  base  the 
line  of  that  base  bisects  the  line  joining  their  vertices. 

190.  A  triangle  X  is  equal  to  a  fixed  triangle  T  and  has  a  common 
base  with  T ';  what  is  the  locus  of  the  vertex  of  X  ?  (Is  the  locus  a  single 
line  or  a  pair  of  lines  ?) 

191.  P  is  any  point  on  the  diagonal  BB  of  A7  ABCB.  Prove  that 
APAB  =  APBC. 

192.  In  ex.  191,  suppose  P  moves  to  B ;  moves  through  D  on  BB 
produced. 

193.  The  sides  AB,  CA  of  a  triangle  are  bisected  in  C ,  B ',  respectively ; 
CC  cuts  BB'  at  P.     Prove  that  A  PBC  =  quadrilateral  AC'PB'. 

194.  If  P  is  a  point  on  the  side  AB,  and  Q  a  point  on  the  opposite 
side  CD  of  O  ABCD,  prove  that  A  PCD  =  A  QA  11. 

195.  If  the  mid-points  of  the  sides  of  any  convex  quadrilateral  are 
joined,  in  order,  then  (1)  a  parallelogram  is  formed.  (2)  which  equals 
half  the  quadrilateral. 


ii 


96 


PLANE    GEOMETRY. 


[Bk.  II. 


Proposition  Y. 

151.  Theorem.  The  rectangle  of  two  given  lines  equals  the 
sum  of  the  rectangles  contained  by  one  of  them  and  the  several 
segments  into  which  the  other  is  divided. 


ax      a2 

X 

ay 

y 

a3 

V 

b 

Given         the  rectangle  of  ax  and  b,  and  b  divided  into  the  seg- 
ments x  and  y. 

To  prove    that  a-J)  =  axx  +  axy. 

Proof.    1.  Let  a2  be  drawn  II  at  from  the  division  point  of  b. 

2.  Then,  in  the  figure,  ax  =  a2  =  as.         I,  prop.  XXIV 

3.  .*.  axb  =  axx  +  axy.  Ax.  8 

Corollaries.  1.  If  a  line  is  divided  internally  into  two 
segments,  the  rectangle  of  the  whole  line  and  one  segment  equals 
the  square  on  that  segment  together  with  the  rectangle  of  the 
two  segments. 

Make  ci\  =  x  in  the  proof  above,  and  consider  step  3. 

2.  If  a  line  is  divided  internally  into  two  segments,  the  square 
on  the  whole  line  equals  the  sum  of  the  rectangles  of  the  whole 
line  and  each  of  the  segments. 

Make  a-i  =  x  +  y  in  the  proof  above. 

Note.  This  theorem  is  the  geometric  form  of  the  Distributive  Law  of 
Multiplication  of  Algebra,  which  asserts  that  a  (x  +  y)  =  ax  +  «2/. 


Exercise.  196.  The  rectangle  of  one  line  and  the  sum  of  two  others 
equals  the  sum  of  the  rectangles  of  the  first  and  each  of  the  other  two. 
Consider  the  case  where  one  of  the  second  two  lines  is  zero  ;  negative. 


SBC.  152.] 


EQUALITY   OF   POLYGONS. 


97 


152.  Positive  and  Negative  Polygons.  In  general,  a  line  AB 
is  thought  of  as  positive ;  but  if,  in  the  discussion  of  a  propo- 
sition, A  is  thought  of  as  approaching  B,  then,  when  A  reaches 
B,  AB  becomes  zero ;  and  if  A  is  thought  of  as  passing  through 
B,  then  AB  is  considered  as  having  passed  through  zero  and 
become  negative ;  that  is,  BA  =  —  AB. 

A  similar  agreement  exists  as  to  triangles. 
In  general,  A  ABC  is  thought  of  as  posi- 
tive ;  but  if,  in  the  discussion  of  a  propo- 
sition, C  moves  down  to  rest  on  AB,  then 
A  ABC  becomes  zero;  and  as  C  passes 
through  AB,  A  ABC  passes  through  zero 
and  is  considered  as  havin 
negative ;  that  is,  A  ACB  = 

In  Book  I,  to  accustom  students  to  this  convention  (that  A  ACB 
=  —  A  ABC),  triangles  were  always  named  by  taking  the  letters  in  the 
counter-clockwise  (or  positive)  order,  except  in  a  few  cases  where  a 
departure  from  this  rule  seemed  advisable. 

A  similar  agreement  exists  as  to  rectangles,  which  illustrates 
the  law  of  signs  in  algebra.  In  the  figure,  I  has  for  its  alti- 
tude and  base  -f  a  and  +  b*  and 
the  rectangle  is  spoken  of  as  4-  ah. 
But  if  h  shrinks  to  zero,  +  ah  also 
shrinks   to   zero,  and  as  h  passes 


changed    its  sign  and  become 
A  ABC. 


II 

-ab4 
-b 


through  zero   and    becomes    nega-    X^ — 
tive,   so  ah  is  considered   to    pass  +ab  -|a 

through  zero  and  to  become  nega- 
tive :  that  is,  II  =  —  ah.  If,  now, 
a  shrinks  to  zero,  and  passes 
through  zero,  changing  its  sign,  so  does  —  ah ;  that  is,  III 
=  +  ah.  And  finally,  as  —  h  again  passes  through  zero,  so 
does  ah,  and  therefore  IV  =  —  ah. 


r 


a  +ab 

+b 


Exercise.     197.    If   P  is  any   point  in  the  plane  of  A  ABC,  then 
A  PAB  -f  A  PBC  +  A  PC  A  =AA  BC.     (Monge.) 


98 


PLANE    GEOMETRY 


[Bk.  II. 


In  the  case  of  polygons  in  general,  the  law  of  signs  will  be  readily 
understood  from  the  annexed  figures.  In  Figs.  1,2,  3  both  the  upper 
and  lower  parts  of  the  polygon  are  considered  as  positive  ;  in  Fig.  4.  P 


pfi 

4 

?, 

C 

0 

C 

y 

/A 

Aj 

A          B 
Fig.  2. 

A\ 

/  + 

/A 

A          B 
Fig.  1. 

A           B 
Fig.  3. 

A 
Fig.  4 

B 

L 

A             B 

Pig.  5. 

has  reached  BC  and  the  upper  part  of  the  polygon  has  become  zero  ;  in 
Fig.  5,  P  has  passed  through  BC  and  the  upper  part  of  the  figure  has 
passed  through  zero  and  become  negative. 

This  treatment  of  negative  surfaces  dates  from  Meister  (1769). 


Propositiox  VI. 

153.    Theorem.      The  square  on  the  sum  of  two  lines  equals 
the  sum  of  the  squares  on  those  lines  plus  twice  their  rectangle. 


n 

I 

H 

y 

xy 

y 

r 

X 

p  y 

X 

X' 

x    xy 

X 

1  y 

A  E        B 

Given  ABCD,  the  square  on  x  +  y. 

To  prove    that  (x  +  y)2  =  x2  +  y2  +  2  xy. 

Proof.    1.   In   the    figure,    let    AE  =  AH=x,    EB  =  HD  =  y, 
EG  WAD,   HFWAB,  and  HE  cut   EG  at  P. 

2.  Then  the  aj's  in  the  figure  arc  all  equal;  also  the  //s. 

Def.  D;  I,  prop.  XXIV 

3.  .'.  AP  =  x%         PC  =  y\  §  99,  def.  □ 
and              EF  =  xy,       HG  =  xy.           §  99,  def.  □ 

4.  /.  (x  +  //)*  -  j-2  +  //  +  2  xy.  Ax.  8 


Prop.  VI. 


EQUALITY   OF  POLYGONS. 


99 


Corollaries.     1.   The  square  on   a  line  equals  four  times 
the  square  on  half  that  Vine. 

Make  x  —  y  in  step  4. 

Then  (2  a;)2  =  .r2  +  x2  +  2  x2, 

or  (2xf  =  4x2. 

That  is.  if  2x  is  the  line,  the  square  upon  it  equals  four 
times  the  square  on  x. 

2.    The  square  on  the  difference  of  two  lines  equals  the  sum 
of  the  squares  on  those  lines  minus  twice  their  rectangle. 


D         Xy 

(x-y>2 
x-y 

C 

r 

A                B 

JL 


For,    in    the    above    figure,    AP  =  x2.    BH  =  y2,   PD  =  xy, 
CM=xy,  and  AC  =  (x  -  y)2. 


But 


A  C  =  AP  +  BH  -  PD-  CH, 


yy 


+  //-  -  2  xy. 


The  truth  of  the  corollary  is,  however,  evident  from  prop.  VI,  if  the 
agreement  as  to  signs  is  considered  ;  for  if  y  becomes  0,  then  2  xy  =  0, 
and  y2  =  0  ;  and  as  y  passes  through  0  and  becomes  negative,  2xy  also 
becomes  negative,  but  y2  remains  positive  because  it  is  the  rectangle  of 
-  y  and  -  y. 


Exercises.     198.    If  ABC MNA  is  the  perimeter  of  any  polygon, 

and  P  is  any  point  in  the  plane,  then  A  PAB  +  APBC  + +  A  PMN 

+  A  PNA  is  constant. 

199.  If  A,  B.  C,  I)  are  four  collinear  points,  in  order,  then  A B  •  CI) 
-f  AD  •  BC  =  AC  •  Bl).  (Euler.)  Investigate  when  B  moves  to  and 
through  C. 


100 


PLANE    GEOMETRY. 


[Bk.  II. 


Proposition  VII. 

154.    Theorem.      The  difference  of  the  squares  on  two  lines 
equals  the  rectangle  of  the  sum  and  difference  of  those  lines. 


x-y     xO» 

F 

y     y2  y 
y 

Given         ABCD,  a  square  on  x,  and  AEFG,  a  square  on  y. 


To  prove    that 


!/2  =  (x  +  U)  (x  -  y)- 


Proof.    1.  Suppose  the  squares  placed  as  in  the  figure,  and  GF 
produced  to  BC.     Then  the  fs  in  the  figure  are  all 

Def.  □ 


equal,  as  also  the  sides  of  x2. 
2.  .'.  the  (x  —  y)'s  are  equal. 


3.  But 

or 

4.  .-.x2 


x-y* 

X' 


Ax.  3 

Ax.  8 

Prop.  V 

Ax.  3 


f2  +  x(x-y)+  y(x-y), 

y2  +  (x  +  tj)(x-y). 

f=        (x+y)(x-y)- 

Corollary.  If  a  point  divides  a  line  internally  or  externally 
into  two  segments,  the  rectangle  of  the  segments  equals  the  differ- 
ence of  the  square  on  half  the  line  and  the  square  on  the  seg^ 
ment  between  the  mid-point  of  the  line  and  the  point  of  division. 

1.  If  AB  is  the  line,  P  (either  Px  or  P2)  the  point  of  divi- 
sion, and  M  the  mid-point,  it  is  to 

be  proved  that  1— , , 

A  p         B  Pa 


AP  ■  PB  =  AM2  -  MP2. 


2.  Let  AB  =  y,  AP  =  x,  then  PB  =  y  -  x,  AM =  \  y,  and 

MP  =  x-\y.   ' 


3.   But  x(y 


->=(!)•-( 


,  by  prop.  VII. 


Sec.  155.] 


EQUALITY   OF  P0LY90XS. 


101 


155.  Reciprocity  between  Algebra  and  Geometry.  From  props. 
V.  VI,  VII,  it  is  evident  that  a  reciprocity  exists  between 
algebra  and  geometry  which  is  likely  to  be  of  great  advantage 
to  each.  This  reciprocity  will  be  more  clearly  seen  by  resort- 
ing to  parallel  columns. 


Geometric  Theorems. 

If  x,  y, are  line-segments, 

and  xy,  xz,  represent  the 

rectangles  of  x  and  y,  x  and 

z, ,  and  x(y-\-z)  represents 

the  rectangle  of  x  and  y  +  z, 
and  x2  represents  the  square 
on  x,  then 

1.  x(y  +  z)=  xy-\-xz. 

Prop.  V 

2.  (x  +  yf=x2  +  y2  +  2xy. 

Prop.  VI 


Algebraic  Theorems. 

If    a,  b,   are    numbers, 

and  ab,   ac represent   the 

products  of  a  and  b,  a  and  r, 

,   and   a(b  +  e)    represents 

the  product  of  a  and  b  +  e, 
and  a2  represents  the  second 
power  of  a,  then 

1.  a  (b  +  e)  =  ab  +  ac. 

2.  (a  +  by-  =  «2  +  b2  +  2ab. 


3. 


x2  =  ± 

Prop.  VI,  cor.  1 

4.  (x  —  y)2  =  x2  -f  y2  —  2 xy. 

Prop.  VI,  cor.  2 

5.  x2  —  y2  =(x-\-y)(x  —  y) . 

Prop.  VII 


"-=<!)' 


4.  (a-by=a?  +  b2-2ab. 

5.  a2-b2  =  (a+b)(a-b). 


This  correspondence  of  one  symbol,  one  operation,  one 
result,  etc.,  of  algebra,  to  one  symbol,  one  operation,  one 
result,  etc.,  of  geometry,  or,  as  it  is  called,  this  "one-to-one 
correspondence"  suggests  many  theorems  jof  geometry  that 
might  otherwise  remain  unnoticed.  This  correspondence  is 
the  basis  of  the  treatment  of  Proportion,  Book  IV. 


Exercise.     200.    Prove  geometrically  that  (x  -f  y)'2  —  (x  —  y)2  =  4xy. 


*l 


102 


PLANE    GEOMETRY. 


[Bk.  II. 


Proposition  VIII. 

156.  Theorem.  In  a  riglit-angled  triangle  the  square  on 
the  hypotenuse  equals  the  sum  of  the  squares  on  the  other 
two  sides. 

,K 


suppose 


E  F    G 

Given         the  right-angled  A  ABC,  Z  C  being  right. 

To  prove    that  a2  +  b2  =  <r. 

Proof.    1.  Let  BLMC  =  a2,  ACKH =  b2,  AEGB  =  c2 
CF  II  AE,  and  HB  and  CE  drawn. 

2.  Then  v  AKCA,  ACB,  are  right,  their  sum  =  st.  Z, 
and  .'.  BCK  is  a  st.  line.  §  14,  def.  st.  Z 

3.  And  v  Z  CAH  =  Z  EAB,  Prel.  prop.  I 
.\  Z  BAH=ZEAC,  by  adding  Z  BAC.  Ax.  2 

4.  And        v  AC  =  #11,  and  ^^  =  £J,       §  99,  def.  □ 

.  • .  A  ABH^  A  ^^C.  Why  ? 

DJ^=  twice  AAEC, 
b2  =  twice  A  J7?jy. 
.'.□  AF,  or  ABAD,  =  b2. 

6.  Similarly,     □  7?^,  or  7L4  •  BD,  =  a2. 

7.  .'.  a2  +  b2=[3AF        +  □  £^=  c2. 


5.  But 

and 


Why? 
Ax.  6 

Axs.  2,  8 


Sec.  157.] 


EQUALITY    OF   POLYGONS. 


103 


157.  Note.  The  first  proof  of  this  theorem  is  said  to  have  been 
given  by  Pythagoras  about  540  b.c,  although  the  theorem  itself  was 
known  long  before  that  time.  From  this  fact  it  is  generally  known  as 
the  Pythagorean  Proposition.    It  is  one  of  the  most  important  in  geometry. 

There  have  been  many  proofs  devised  for  the  Pythagorean 
proposition.  In  the  subsequent  exercises  occasional  proofs 
will  be  suggested,  that  the  student  may  see  the  great  variety 
of  ways  in  which  the  theorem  may  be  attacked.  That  the 
proposition  would  naturally  be  suggested  to  a  people  using 
tile  floors  is  seen  from  Fig.  1,  although  the  proof  following 


Fig.  2. 


Fig.  1. 


from  such  a  figure  is  special,  being  limited  to  the  case  of  the 
isosceles  right-angled  triangle. 

In  Fig.  2  is  given  a  suggestion  of  the  conjectured  proof  of 
Pythagoras :  If  A  1,  2,  3,  4  are  taken  from  the  figure,  the 
square  on  the  hypotenuse  remains ;  and  if  the  two  CD  AP, 
PB,  are  taken  away,  the  sum  of  the  squares  on  the  two  sides 
remains  ;  but  since  the  two  rectangles  equal  the  four  triangles, 
these  remainders  are  equal. 


Exercises.     201.    What  is  the  use  of   steps  2  and  3  in  the  proof  of 
prop.  VIII  ? 

202.  In  the  figure  of  prop.  VIII,  prove  that  AK  II  BM. 

203.  Also  that  H,  C,  L  are  collinear. 

204.  Twice  the  sum  of  the  squares  on  the  medians  of  a  right-angled 
triangle  equals  thrice  the  square  on  the  hypotenuse. 


104 


PLANE    GEOMETRY. 


[Bk.  II. 


Fig.  3  is  that  of  Bhaskara,  the  Hindu :  The  inside  square 
is  evidently  (a  —  b)2,  and  each  of  the  four  triangles  is  \  ab ; 


.'.  & 


4  .\ ab  =  (a  -  bf  =  a2  +  b2  -  2 ab  ;  .'.  c2  =  a2  +  b\ 


a  / 

^  A 


Fig.  3. 


Fig.  4. 


Fig.  4  is  one  of  the  most  simple :  If  from  the  whole  figure 
there  are  taken  A  b,  there  remains  the  square  on  the  hypote- 
nuse ;  or  if  the  equal  A  a  are  taken,  there  remains  the  sum  of 
the  squares  on  the  two  sides. 

158.  Definition.  The  projection  of  a  point  on  a  line  is  the 
foot  of  the  perpendicular  from  the  point  to  the  line. 

Thus  A'  and  B',  Figs.  1,2,  are  the  projections  of  A  and  B  on  X'X. 

The  projection  of  a  line-segment  on  another  line  in  the  same 
plane  is  the  segment  cut  off  by  the  projections  of  its  end- 
points,  e.g.  in  Figs.  1  and  2,  A'B'  is  the  projection  of  AB. 


A 

A 

n             *w 

p 

\ 

^^.B 

\      r- 

"N5  r 

Ab 

(' 

X       B'       \ 

\  f 

\ 

A'  B' 

Fig.  1. 


NT 


Fig.  2. 


A'  B'   A'         B' 

Fig.  3.  Fig.  4. 


Strictly  these  are  orthogonal  (or  right-angled)  projections ;  but  since 
orthogonal  projections  are  the  only  kind  ordinarily  considered  in  elemen- 
tary geometry,  they  are  called,  simply,  projections.  In  advanced  geom- 
etry, the  projections  of  Figs.  3,  4  are  among  the  others  used.  Fig.  3 
represents  an  oblique  projection,  and  Fig.  4  represents  a  projection  from 
a  point.  Fig.  4  is  the  most  general,  approaching  the  others  as  P  recedes 
to  a  greater  distance. 


Prop.  IX. 


EQUALITY   OF  POLYGONS. 


105 


Proposition  IX. 


159.  Theorem.  In  an  obtuse-angled  triangle  the  square  on 
the  side  opposite  the  obtuse  angle  equals  the  sum  of  the 
squares  on  the  other  two  sides,  together  with  twice  the  rect- 
angle of  either  side  and  the  projection  of  the  other  on  the 
line  of  that  side. 


Given         A  abc,  obtuse-angled  opposite  c,  and  a'  the  projection 
of  a  on  the  line  of  b. 

To  prove    that  c2  =  a2  +  b2  -f  2  ba '. 

Proof.    1.  In  the  figure,  h  _L  a',  §  158,  clef,  projection 

.\h2-\-(a'+b)2  =  c2,  Why? 

or  h2  +  a'2  +  b2  +  2  a'b  =  c\  Prop.  VI 


.-.  a2+b2  +  2a'b 


Prop.  VIII 


Exercises.  205.  In  the  figure  of  prop.  VIII,  prove  that,  if  II K  and 
LMaxe  produced  to  meet  at  P,  then  AE  =  and  II  PC,  and  BG  =  and  II  PC. 

206.  If  the  diagonals  of  a  quadrilateral  intersect  at  right  angles,  prove 
that  the  sum  of  the  squares  on  one  pair  of  opposite  sides  equals  the  sum 
of  the  squares  on  the  other  pair. 

207.  In  the  annexed  figure,  equilateral  triangles 
are  constructed  on  the  sides  of  a  right-angled 
triangle;  M  is  the  mid-point  of  CA.  Prove  (1) 
A  ABK  ^  A  AEC\  (2)  MK  II  BC,  (3)  A  BCM  = 
A  BCK,  (4)  A  BRM  =  A  RCK,  (5)  A  ABE  = 
AACK  +  A  ABM  =  A  ACE  +  iAABC,  (6) 
.-.  from  (1)  and  (5)  AAEC  =  A  ACE  +  i  A  ABC. 
(7)  similarly,  A  CEB  =  A  BLC  +  £  A  ABC,  (8) 
.-.  figure  AEBC  =  figure  ABLCE,  (9)  .-.  A  AEB 
=  A  BLC  +  A  ACE.  State  in  full  form  the 
theorem  proved  in  (9). 


106  PLANE    GEOMETRY.  [Bk.  II. 

Corollaries.  1.  In  any  triangle  the  square  on  the  side 
opposite  an  acute  angle  equals  the  sum  of  the  squares  on  the 
other  two  sides,  less  twice  the  rectangle  of  either  side  and  the 
projection  of  the  other  side  on  it. 


b 

For,  in  the  above  figure, 

h°~  +  (b-  a')2  =  c\ 

.-.  h2  +  b2  +  a'*-2a'b  =  c*. 

.-.  a2  +  b*-2a'b  =  c2. 

The  truth  of  the  corollary  is,  however,  evident  from  prop.  IX  ;  for  if 
Z  ba  becomes  90°,  a'  =  0  and  prop.  IX  becomes  prop.  VIII  ;  and  if  Z  b<i 
becomes  acute,  a'  passes  through  0  and  becomes  negative,  and  □  a'b 
becomes  negative  ;  .-.  step  2  becomes  a2  +  b'2  —  2  a'b  =  c2. 

2.  Converse  of  props.  VIII,  IX,  and  prop.  IX,  cor.  1.  The 
angle  opposite  a  given  side  of  a  triangle  is  right,  obtuse,  or 
acute,  according  as  the  square  on  that  side  is  equal  to,  greater 
or  less  than  the  sum  of  the  squares  on  the  other  two  sides. 

Law  of  Converse  (§  73).     Write  out  the  proof  in  full. 


Exercises.  208.  In  the  figure  of  prop.  VIII,  the  medians  of  A  ABC 
are  perpendicular  to  and  equal  to  half  of  KM,  HE,  LG,  respectively. 
(Complete  the  \3  BCAV,  and  prove  CV  —  and  JL  KM,  etc.) 

209.  XOY  is  any  angle,  and  from  B,  on  OY.  BA  is  drawn  JL  OX  ; 
from  B  is  drawn  BZ  II  OX ;  now  if  P  can  be  found  on  BA,  so  that  OP 
produced  to  cut  BZ  in  Q,  makes  PQ  =  2  OB,  then  Z  XOQ  =  $  Z  XOY. 
(That  is,  A  XOY  is  trisected.  It  has  been  proved  that  this  famous 
problem  of  the  Greeks,  to  trisect  any  angle,  cannot  be  solved  by 
elementary  geometry,  that  is,  by  using  the  compasses  and  straight-edge 
only.     There  are  various  solutions  if  other  instruments  are  allowed.) 

210.  Prove  algebraically  that  if  n  is  an  even  number,  then  n,  \  n2  —  1, 
-}n2  +  1  are  numerically  the  sides  of  a  right-angled  triangle  (Plato),  and 
that  they  are  integers. 


Prop,  X.]  EQUALITY    OF   POLYGONS.  107 

Pboposition  X. 

160.  Theorem.  The  sum  of  the  squares  on  any  two  sides  of 
a  triangle  equals  twice  the  sum  of  the  squares  on  one-half  the 
third  side  and  on  the  median  to  that  side. 


c 
Given         the  A  abc.  and  m  the  .median  to  c. 


To  prove    that         <<-  +  b2 


[(i)"-] 


Proof.    1.  Let  mJ  be   the   projection  of  m   on  c  and  suppose 
Z  cm  acute. 

2.   Then  a1  =(  |  J   +  m?  -  2  (  |  J  ///,      Prop.  IX,  cor.  1 


? 


and      /,-  =    -      -f  „r  +  21-     ///'.  Why 


..«.+if=2  [(!)"+»-] 


Ax.  2 


If  Z  cw  is  obtuse,  then  Z  ?ytc  is  acute,  and  the  proof 
merely  interchanges  a.  b  without  affecting  step  3. 
If  Z  cm  is  right,  then  m'  =  0  in  step  2,  but  3  is  not 
affected. 

Exercises.  211.  In  prop.  X.  prove  that  4  m2  =  2  (a2  +  b2)  -  c2.  Hence 
show  that  in  a  right-angled  triangle  (in  which  a2  -f  fr2  =  c2)  the  median 
to  the  hypotenuse  equals  half  the  hypotenuse. 

212.  From  ex.  211,  what  is  the  locus  of  the  vertex  of  the  right  angle 
of  a  right-angled  triangle  with  a  given  hypotenuse  ? 

213.  The  sides  of  a  triangle  are  10.  12,  15  inches.  Is  the  triangle 
right-angled  ?    ohtuse-an^led  ? 


V 


\ 


108  PLANE   GEOMETRY.  [Bk.  II. 

Proposition  XI. 

161.  Theorem.  The  sum  of  the  squares  on  the  sides  of  a 
quadrilateral  equals  the  sum  of  the  squares  on  the  diagonals 
plus  four  times  the  square  on  the  line  joining  the  mid-points 
of  the  diagonals. 

Pr 


Given  a  quadrilateral  abed,  convex,  concave,  or  cross,  with 
diagonals  e,  f  and  with  m  joining  the  mid-points 
of  e,  f. 

To  prove    that  a2  +  b2  +  c2  +  d2  =  e2  +  f2  +  4  m2. 


Proof.    1.  In  the  figure,  a2  +  d2  =  2  x2  +  2  I  %  J  ,  Why  ? 


•GO- 


? 


00 


? 


and  b2  +  c2  =  2y2  +  2  (%     •  Why? 


2.  .'.  a2  +  62  +  c2  +  d2  =  2  (x2  +  y2)  +  4  (  {  )  Ax.  2 

2 


Prop.  X 
=  e2  +/2  +  4  m2.  Prop.  VI,  cor.  1 


Corollary.      The  sum  of  the  squares  on  the  diagonals  of  a 
parallelogram  equals  the  sum  of  the  squares  on  the  sides. 
For  then  m  =  0;  I,  prop.  XXIV,  cor.  2. 

Note.     The  theorem  is  due  to  Euler.     The  corollary  was,  however, 
known  to  the  Greeks. 


^ 


Prop.  XII.] 


PROBLEMS. 


109 


2.     PROBLEMS. 
Proposition  XII. 

162.    Problem.      To  construct  a  triangle  equal   to   a  given 
poly g  o)i. 

D  • 


Fig.  2. 

Given         polygon  ABCDE. 

Required    to  construct  a  A  equal  to  ABCDE. 

Construction.    1.  Produce  BA,  join  D  and  A,  draw  EF  II  DA, 
meeting  BA  produced  at  F ;  draw  DF. 

§  28,  post,  of  st.  line ;  I,  prop.  XXXIII 

2.  Then  polygon  FBCD  has  one  less  side  than  ABCDE, 
and  will  be  proved  equal  to  it.  Continue  the  process 
until  a  A  is  reached  (Fig.  2). 

Proof.    1.  v  EFW  DA, 

.-.A  ADF  =  A  ADE,  having  same  base  AD. 

Prop.  IT 

2.  Adding  polygon  ABCD,  FBCD  =  ABCDE.      Ax.  2 

3.  Similarly  thereafter.  In  Pig.  2.  A  FGD  is  the  tri- 
angle required. 


Exercise.     214.    To  construct  a  rhombus  equal  to  a  given  parallelo- 
gram, and  on  the  same  base.     Discuss  for  impossible  cases. 


110 


PLANE    GEOMETRY. 


[Bk.  II. 


Proposition  XIII. 

163.    Problem.      To   construct   a   square    equal   to   a  given 
polygon. 

^'— -E h 


\S 


D  E    G 
C 


Prop.  XII 


Given         polygon  G. 

Required    to  construct  a  square  equal  to  G. 

Construction.    1.  Construct  a  A  equal  to  G. 

2.  By  drawing  a  line  through  the  vertex  of  this  A  II  to 
the  base,  and  erecting  _L's  from  an  extremity  and 
the  mid-point  of  the  base,  construct  a  □,  as  ABCD, 
equal  to  this  A.  I,  props.  XXIX,  XXXI,  XXXIII 
Then  if  AB  ==  DA,  ABCD  is  the  required  □. 

3.  If  not,  produce  AD  to  E,  making  DE  =  CD;  §  28 
bisect  AE  at  0,  I,  prop.  XXXI 
and  with  center  0  and  radius  OE,  describe  a  semi- 
circumference.                                        §  109,  post,  of  O 

4.  Produce  CD  to  meet  circumference  at  F,  §  28 
and  construct  a  square  on  DF.  I,  prop.  XXX IX 
Then  DF2,  S  in  the  figure,  is  the  required  □. 

Proof.    1.  Draw  OF,  let  r  =  OF  =  OA  =  OE,  and  x  =  OD\ 
then  CD  =  DF  =  r  —  x, 

and  AD  =  r  +  x. 

2.  Then  (r  -f  x)  (r  -  a-)  =  r1  -  x2  Prop.  Vil 


+  DF' 


•a  =  &F*.   Why? 


3.    Hut    (r  +  x)(r  —  x)  =U}ABCD 
and  .*.  DF2  =  polygon  < > 


<;. 


Const.  2 
Ax.  1 


Exs.  215-2:30.]  PROBLEMS.  HI 

EXERCISES. 

215.  If  one  angle  of  a  triangle  is  two-thirds  of  a  straight  angle,  show 
that  the  square  on  the  opposite  side  equals  the  sum  of  the  squares  on  the 
other  two  sides,  together  with  their  rectangle. 

216.  Prove  prop.  XI  for  a  concave  quadrilateral. 

217.  If  ZP  =  180°  and  SP  =  PQ.  show  that  prop.  XI  reduces  to  a 
previous  theorem. 

218.  Prove  prop.  XI.  cor.  directly  from  prop.  X  without  reference  to      "% 
prop.  XL 

219.  If  ABCD  is  any  quadrilateral,  and  the  mid-points  of  the  diagonals 
^  "are  joined  by  a  line  bisected  at  -If,  and  if  P  is  any  point,  then  PA2  +  PB2 

+  PC2  -  PD-  =  MA*  +  MB2  +  M&  +  ML2  +  4  PM*. 

220.  To  construct  a  parallelogram  equal 

to  a  given  triangle,  and  having  one  of  its       C D E 

"angles  equal  to  a  given  angle.  V^v          /           / 

221.  To  construct  a  parallelogram  equal  \  \^s/  /  / 
to  a  given  square,  on  the  same  base  and  V___y_^Ny  Xn 
having  an  angle  equal  to  half  the  angle  of  A         M           B 

the  square. 

222.  To  construct  an  isosceles  triangle  equal  to  a  given  triangle,  and 
on  the  same  base. 

223.  To  construct  a  triangle  equal  to  a  given  parallelogram,  and  having 
one  of  its  angles  equal  to  a  given  angle. 

224.  To  construct  a  parallelogram  equal  to  a  given  triangle,  and  having 
its  perimeter  equal  to  that  of  the  triangle.  (In  the  figure  of  ex.  220  how 
must  3ID  compare  with  BC  +  CA  ?) 

225.  To  construct  a  square  equal  to  the  sum  of  two  given  squares. 
(Apply  prop.  VIII.) 

226.  On  a  given  line  to  construct  a  rectangle  equal  to  a  given  rectangle. 

227.  On  one  side  of  a  triangle  as  a  diagonal  to  construct  a  rhombus 
equal  to  the  given  triangle. 

228.  Prove  that  in  any  triangle  three  times  the  sum  of  the  squares  on 
the  sides  equals  four  times  the  sum  of  the  squares  on  the  three  medians. 

229.  Also  that  three  times  the  sum  of  the  squares  on  the  lines  joining 
the  centroid  to  the  vertices  equals  the  sum  of  the  squares  on  the  sides. 

230.  If  one  angle  of  a  triangle  is  one-third  of  a  straight  angle,  show 
that  the  square  on  the  opposite  side  equals  the  sum  of  the  squares  on  the 
other  two  sides  less  their  rectangle. 


112  PLANE    GEOMETRY.  [Bk.  II, 


3.     PRACTICAL   MENSURATION. 

164.  For  practical  purposes  a  surface  is  measured  as 
follows  : 

1.  A  square  unit  is  defined  as  a  square  which  is  one  linear 
unit  long  and  one  linear  unit  wide. 

That  is,  a  square  inch  is  a  square  that  is  1  in.  long  and  1  in.  wide  ;  a 
square  meter  is  a  square  that  is  1  m.  long  and  1  m.  wide,  etc.  In  the 
figure  the  shaded  square  is  considered  as  a  square  unit. 

2.  If  two  sides  of  a  rectangle  are  3  in.  and  5  in.  respec- 
tively, then,  in  the  figure,  the   area  of  the 

strip  AB  is  5  X  1  sq.  in.,  and  the  total  area 

is  3  X  5  X  1  sq.  in.,  or  15  sq.  in.  ^ B 

Theoretically,  a  rectangle  rarely  has  sides  Jj~ — — — — 
both    of   which    exactly   contain    any   linear 
unit,  however  small.     Such  cases  are  discussed  in  Book  IV. 

But  for  practical  purposes  the  above  method  is  approximate 
to  any  required  degree. 

At  present  it  is  necessary  for  the  student  to  learn  that  geometry  gives 
him  an  instrument  for  practical  work.  It  will  accordingly  be  assumed 
that  the  measurements  can  be  made  to  any  degree  of  approximation,  and 
that  the  expressions  area,  measure,  etc. ,  are  understood  in  their  ordinary 
sense.  It  has  already  been  explained  that  the  rectangle  of  two  lines 
corresponds  to  the  product  of  two  numbers ;  hence,  in  practice,  lines  ar<J 
represented  by  numbers,  and  their  rectangles  by  the  products  of  those 
numbers.  This  practical  measurement  will  be  exemplified  hereafter,  as 
it  has  already  been  to  some  extent,  in  the  numerical  exercises. 


Exercises.  231.  A  field  is  in  the  form  of  a  rhombus,  the  obtuse  angle 
being  twice  the  acute  angle ;  the  shorter  diagonal  is  300  feet.  Find  the 
area  of  the  field  in  square  feet. 

232.  A  railroad  embankment  extends  through  a  farm  1  mile  long,  its 
rails  being  in  straight  lines  perpendicular  to  the  two  parallel  sides  of  the 
farm  ;  the  embankment  is  80  ft.  wide  at  the  bottom  at  one  end,  and  00  ft. 
at  the  other.      How  much  land  was  taken  for  railroad  purposes? 


Exs.  233-247.]  PRACTICAL    MENSURATION.  Hi 


EXERCISES. 

233.  A  road  running  across  a  farm  is  i  mile  long  and  3  rods  wide; 
the  road  being  rectangular,  find  its  area  in  acres. 

234.  The  side  of  an  equilateral  triangle  is  15.     Find  the  area. 

235.  In  excavating  for  a  canal  30  ft.  deep,  200  ft.  wide  at  the  top,  and 
160  ft.  wide  at  the  bottom,  what  is  the  area  of  a  cross-section  '? 

236.  One  diagonal  of  a  quadrilateral  is  100.  and  the  perpendiculars, 
from  the  other  two  vertices,  upon  it,  are  50  and  40.     Find  the  area. 

237.  The  area  of  a  triangle  is  a  and  the  altitude  is  h.  Find  the  base. 
Investigate  for  a  =  325.85,  h  =  38  ;  also  for  a  =  100,  h  —  100. 

238.  The  area  of  a  trapezoid  is  a  and  the  two  bases  are  bi,  b2.  Find 
the  altitude.  Investigate  for  a  =  223.375.  &i  =  13.5,  62  =  6.4  ;  also  for 
a  =  10.  &!  =  0.  b2  =  10. 

239.  The  area  of  a  trapezoid  is  542.5.  the  altitude  is  21.7,  and  the 
difference  between  the  parallel  sides  is  11.2.     Find  those  sides. 

240.  The  area  of  a  square  is  2.  Find  the  side  of  a  square  of  twice  the 
area ;  thrice  the  area  ;  four  times  the  area. 

241.  The  altitude  of  an  equilateral  triangle  is  160.     Find  the  area. 

242.  The  base  of  an  isosceles  triangle  is  f  of  one  of  the  equal  sides, 
and  the  altitude  is  10.     Find  the  area. 

243.  Two  sides  of  a  right-angled  triangle  are  1036  and  1173.  Find 
the  hypotenuse  and  the  area. 

244.  Find  to  three  decimal  places  the  diagonal  of  a  square  whose  area 
is  1. 

245.  In  a  right-angled  triangle  the  perpendicular  from  the  vertex  of 
the  right  angle  divides  the  hypotenuse  into  two  segments,  2.88  and  5.12. 
Find  the  two  sides. 

246.  From  the  vertex  A  of  A  ABC.  AD  ±  BC.  Find  the  lengths  of 
BD.  CD.  knowing  that  AB  =  307.8.  CA  =  480.168,  BC  =  689.472. 

247.  A  surveyor,  wishing  to  erect  a  perpendicular  to  a  line  on  the 
ground,  drives  two  stakes,  A,  B.  12  links  apart;  to 
these  he  fastens  the  ends  of  a  24-link  segment,  and 
stretches  the  chain,  at  the  end  of  the  9th  link  from 
A,  to  C.  Show  that  AC  ±  AB.  (This  method  of 
erecting  perpendiculars  was  known  to  the  temple 
and  pyramid  builders,  and  surveyors  employed  for 
this  purpose  were  called  '-rope  stretchers."  The  method  is  still  used  in 
practical  field  work.) 


BOOK   III.  — CIRCLES. 


165.  Definitions.  A  circle  is  the  finite  portion  of  a  plane 
bounded  by  a  curve,  which  is  called  the  circumference,  and  is 
such  that  all  points  on  that  line  are  equidistant  from  a  point 
within  the  figure  called  the  center  of  the  circle. 

For  corollaries  and  postulates,  see  §§  108,  109. 

Certain  definitions  are  here  repeated  for  convenience. 

If  two  equal  figures  are  necessarily  congruent,  as  in  the  case  of  circles, 
angles,  squares,  and  line-segments,  the  word  equal  is  ordinarily  used  to 
express  congruence.  Hence  congruent  circles  (see  §  108,  2)  are  ordinarily 
•called  simply  equal. 

A  straight  line  terminated  by  the 
center  and  the  circumference  is  called 
a  radius. 

A  straight  line  through  the  center 
terminated  both  ways  by  the  cir- 
cumference is  called  a  diameter. 

166.  The  straight  line  joining  any 

two  points  on  a  circumference  is  called  a  chord. 

Hence  a  diameter  is  a  chord  passing  through  the  center.  In  the 
figure,  AE  and  BB  are  chords. 

The  expressions  center,  radius,  diameter,  chord,  of  a  circumference 
are  sometimes  used  instead  of  center,  etc.,  of  a  circle. 

167.  The  line  of  which  a  chord  is  a  segment  is  called  a 
secant,  as  XY  in  the  figure. 

168.  A  part  of  a  circumference  is  called  an  arc. 

In  the  figure,  BCD  is  an  arc.  As  in  naming  an  angle,  the  counter- 
clockwise order  is  followed,  and  arcs  so  named  are  considered  positive. 

114 


Sacs.  169-177.]  CIRCLES.  115 

169.  One-half  of  a  circumference  is  called  a  semicircum- 
ference. 

170.  A  fourth  part  of  a  circumference  is  called  a  quadrant. 

171.  An  angle  formed  by  two  radii  is  called  a  central  angle. 
In  the  figure,  AAOB,  BOE  are  central  angles. 

172.  A  central  angle  is  said  to  stand  upon  the  arc  which  lies 
within  the  angle  and  is  cut  off  by  the  arms. 

A  AOB.  BOE  stand  upon  AB,  BE,  respectively. 

173.  The  arc  upon  which  the  sum  of  two  central  angles 
stands  is  called  the  sum  of  the  arcs  upon  which  those  angles 
stand.     Similarly  for  the  difference  of  two  arcs. 

Thus,  AE  =  AB  +  BE.  and  AB  =  AD  -  BI). 

174.  Two  arcs  are  said  to  be  complements  of  each  other  if 
their  sum  is  a  quadrant ;  supplements  of  each  other  if  their 
sum  is  a  semicircumference  ;  conjugates  of  each  other  if  their 
sum  is  a  circumference. 

In  the  figure.  AB  is  the  supplement  of  BE  and  the  conjugate  of  BA. 

175.  An  arc  greater  than  a  semicircumference  is  called  a 
major  arc ;  one  less  than  a  semicircumference.  a  minor  arc. 

In  the  figure,  AB,  BI).  DE  are  minor  arcs  ;  BE  A  is  a  major  arc. 

176.  Conjugate  arcs  are  said  t<>  be  subtended  by  their  com- 
mon chord. 

In  the  figure.  BI)  and  I)B  are  each  said  to  be  subtended  by  chord  BI). 

The  word  suhtend  is  variously  used  in  geometry.  It  means  to  extend 
under  or  to  be  opposite  to.  Hence  in  a  triangle  a  side  is  said  to  subtend 
an  opposite  angle,  a  chord  is  said  to  subtend  an  arc,  etc. 

177.  A  portion  of  a  circle  cut  off  by  an  arc  and  two  radii 
drawn  to  its  extremities  is  called  a  sector,  and  the  central 
angle  standing  on  that  arc  is  called  the  angle  of  the  sector. 

In  the  figure.  OAB  is  a  sector,  and  LAOB  is  its  angle. 


116 


PLANE    GEOMETRY. 


[Bk.  III. 


1.     CENTRAL  ANGLES. 
Proposition  I. 

178.  Theorem.  In  the  sa?ne  circle  or  in  equal  circles,  if 
two  central  angles  are  equal,  the  arcs  on  which  they  stand 
are  equal  also,  and  of  two  unequal  central  angles  the  greater 
stands  on  the  greater  arc. 


Given         M,  M',  two  equal  circles,  and  central  angles 
AOB  =  A'O'B',   AOOA'O'B'. 

To  prove    that     AB  =  A^B',  and  AC  >  jVB'. 

Proof.    1.  Place  O  M '  on  O  M  so  that  Z  A'O'B'  coincides  with 
its  equal  Z  A  OB. 
Then  A'  coincides  with  A,  and  B'  with  B. 

§  165,  def.  0 

2.  Then  A'B'  coincides  with  AB,  because  its  points  are 
equidistant  from  0.  §  165,  def.  O 

3.  Also,  v  ZAOOZ  A'O'B', 

.-.ZAOO  ZAOB. 

4.  .*.  C  is  not  in  Z  A  OB,  and  AC  >  IB.  Ax.  8 

5.  And  v  AB  =  ATB',  .'.  AC  >  ArB'.  Ax.  9 
The  proof  is  essentially  the  same  for  a  single  circle, 
and  so  in  general  when  equal  circles  are  involved. 


Prop.  I.]  CENTRAL    ANGLES.  117 

Corollaries.  1.  Sectors  of  the  same  circL  or  of  equal 
circles,   which   have  equal  angles,' are  equal. 

For,  by  steps  1,  2,  they  coincide.. 

2.  Sectors  of  the  same  circle,  or  of  equal  circles,  which  have 
unequal  angles,  are  unequal,  the  greater  having  the  greater 

angle. 

This  is  proved  by  superposition  in  steps  3.  4.  5,  of  the  proposition. 

3.  The  two  arcs   into  which  the  circumference   is  divided  by 

a  diameter  are  equal . 

For  their  central  angles  are  straight  angles,  and  these  being  equal  the 
arcs  are  equal  by  the  proposition. 

4.  The  two  figures  into  which  a  circle  is  divided  by  a  diam- 
eter are  equal. 

For  their  central  angles  are  straight  angles.  Hence  by  cor.  1  they 
are  equal. 

This  corollary  is  attributed  to  Thales. 

179.  Definition.  The  figure  formed  by  a  semicircumference 
and  the  diameter  joining  its  extremities  is  called  a  semicircle. 

It  is  proved  (cor.  4)  that  all  semicircles,  cut  from  the  same  circle,  are 
equal.     Hence  the  name,  semi-  meaning  half. 

180.  Since  the  360  equal  angles,  into  which  the  perigon 
at  the  center  of  a  circle  is  imagined  to  be  divided,  stand  on 
equal  arcs  by  prop.  I,  the  ordinary  mensuration  of  angles  by 
degrees  is  also  used  for  arcs.  Similarly  for  minutes,  seconds, 
and  other  measurements.  Hence  the  common  expression,  an 
angle  at  the  center  is  measured  by  the  subtended  arc. 

The  expression  is  not  strictly  correct ;  we  do  not  measure  an  angle  by 
an  arc,  but  the  angle  and  arc  have  the  same  numerical  measure,  as  will 
be  proved  in  §  254.  We  might  as  truly  say  that  an  arc  is  measured  by 
its  central  angle.  But  the  expression  is  so  commonly  used,  and  has  found 
its  way  into  so  many  text-books  and  examination  papers,  that  the  student 
needs  to  become  familiar  with  it. 


118  PLANE    GEOMETRY.  [Bk.  III. 

Proposition  II. 

181.  Theorem.  In  the  same  circle  or  in  equal  circles,  if 
two  arcs  are  equal,  the  central  angles  which  they  subtend  are 
equal  also,  and  of  two  unequal  arcs  the  greater  subtends  the 
greater  central  angle. 

Proof.  If  0  and  0'  are  two  central  angles,  and  A,  A'  are  the 
arcs  on  which  they  stand,  it  has  been  proved  in  prop.  I 
that 

If   0  >  0',     then     A  >  A', 

«    0  =  0',       "        A  =  A', 

«    0  <  0',       «        A<  A'. 

Hence  the  converses  are  true,  by  the  Law  of  Converse,  §  73. 

Corollaries.  1.  In  the  same  circle  or  in  equal  circles, 
equal  sectors  have  equal  angles;  and  of  two  unequal  sectors, 
the  greater  has  the  greater  angle. 

Law  of  Converse,  §  73,  from  prop.  I,  cors.  1,  2. 

2.  A  central  angle  is  greater  than,  equal  to,  or  less  than,  a 
right  angle,  according  as  the  arc  on  which,  it  stands  is  greater 
than,  equal  to,  or  less  than,  a  quadrant.      (Why  ?) 


Exercises.  248.  If  two  lines  drawn  to  a  circumference,  from  a  point 
within  the  circle,  are  equal,  they  subtend  equal  central  angles. 

249.  Prove  the  converse  of  ex.  248. 

250.  Two  circumferences  cannot  bisect  each  other. 

251.  Suppose  from  the  point  P  on  a  circumference  two  equal  chords, 
PA,  PB,  are  drawn.  Prove  (1)  that  these  chords  subtend  equal  central 
angles,  (2)  that  they  subtend  equal  arcs. 

252.  The  arc  AB  is  bisected  by  the  point  M,  and  MC  is  a  diameter  ; 
prove  that  chord  AC  =  chord  BC. 

253.  How  many  degrees  in  the  central  angle  standing  on  a  third  of  a 
circumference?   a  fourth  ?   a  fifth  ? 


Prop.  III.]  CHORDS   AND    TANGENTS.  119 

2.     CHORDS    AND   TANGENTS. 

Proposition  III. 

182.  Theorem.  In  the  same  circle  or  in  equal  circles,  if 
two  arcs  are  equal  they  are  subtended  by  equal  chords,  and 
of  two  unequal  minor  arcs  the  greater  is  subtended  by  the 
greater  chord. 


Given         two  equal  circles,  M,  M' ;   two  equal  arcs,  K,  K' ; 
and  two  unequal  minor  arcs,  K  >  K". 

To  prove    that,  as  lettered  in  the  figure,  chords  AB  =  A'B', 
AB  >  CA'. 

Proof.    1.  Draw  the  radii  OA,  OB,  O'A',  O'B',  O'C.     Then 

v  K=  K',     .'.  Z  AOB  =  Z  A'O'B'.  Prop.  II 

2.  But  v  OM=OM', 

.-.  OA  =  OB=  O'A'  =  O'B'  =  O'C. 

3.  .'.A  OAB  ^  A  O'A'B',  and  AB  =  A'B'.         Why  ? 

4.  Also, 

V  K>  K",     r.A  AOB  >  Z  CO' A',        Prop.  II 
.-.  AB  >  CA'.  I,  prop.  X 

Corollary.     In  the  same  circle  or  in  equal  circles,  of  two 
unequal  major  arcs,  the  greater  is  subtended  by  the  less  chord. 


120  PLANE    GEOMETRY.  [Bk.  III. 

Proposition  IV. 

183.  Theorem.  In  the  same  circle  or  in  equal  circles,  if 
two  chords  are  equal  they  subtend  equal  major  and  equal 
minor  arcs  ;  and  of  two  unequal  chords  the  greater  subtends 
the  greater  minor  and  the  less  major  arc. 

Proof.     Let  C,  C  be  two  chords  of  the  same  circle  or  of  equal 
circles  ;  JVJ  N'  their  corresponding  minor  arcs  ; 
J}  J'       "  "  major  arcs. 

From  prop.  Ill, 

if  N  >  N',  or  if  J  <  J\  then   C  >  C, 
'<  N=N',    "    «   J=J',      «      C  =  C, 
"  JY<N',    «    «   J>J',      "      C<C. 
Hence  the  converses  are  true,  by  the  Law  of  Converse,  §  73. 


Exercises.  254.  If  through  a  point  in  a  circle  two  chords  are  drawn 
making  equal  angles  with  the  diameter  through  that  point,  these  chords 
cut  off  equal  arcs  of  the  circle. 

255.  The  intersecting  chords  joining  the  extremities  of  two  equal  arcs 
of  a  circle  are  equal. 

256.  What  is  meant"  by  an  arc  of  75°  ?  by  one  of  300°  ?  Can  the  sum 
of  two  arcs  ever  exceed  an  arc  of  300°?  Draw  a  figure  to  illustrate 
your  answer. 

257.  Does  the  chord  subtending  the  arc  2  a  equal  twice  the  chord  sub- 
tending the  arc  a  ?     Prove  your  statement. 

258.  May  the  chord  subtending  the  arc  2  a  ever  equal  the  chord  sub- 
tending the  arc  a  ?  If  not,  show  why  ;  if  so,  tell  how  many  degrees  in 
the  arc  a. 

259.  How  many  degrees  in  the  supplement  of  the  arc  90°?  175°? 
180°?    190°? 

260.  How  many  degrees  in  the  conjugate  of  the  arc  180°?  360°? 
360°?   400°? 

261.  How  does  the  length  of  the  chord  subtending  an  arc  of  60°  com- 
pare with  that  subtending  an  arc  of  90°  ?  300°  ?  (Call  the  radius  r,  and 
determine  each  in  terms  of  r.) 


Prop.  V.]  CHORDS  AND    TANGENTS.  121 

Proposition  V. 

184.    Theorem.     A  diameter  which  is  perpendicular  to  a 
chord  bisects  the  chord  and  its  subtended  arcs. 


Given         the  diameter  BD  perpendicular  to  chord  AC  at  E. 
To  prove    that  (1)  AE  =  EC, 

(2)  AB  =  jfC,  (3)  DA  =  CD. 

Proof.    1.  Drawing  radii  OA,  OC,  then 

OA  =  OC,  §  109,  post,  of  O 

and                     AE  =  EC,  I,  prop.  XX,  cor.  6 

.'.ZAOE  =  Z  EO  C.  I,  prop.  XX,  cor.  5 

2.  .'.AB  =  BC.  Why? 

3.  And        v  Z  DOA  =  Z  COD,  Prel.  prop.  IV 

.-.DA  =  CI).  Why? 

Corollaries.  1.  Conversely,  a  diameter  which  bisects  a 
chord  is  perpendicular  to   it. 

For  v  AE  —  EC,  and  OA  =  00,  .-.  DB  has  two  points  equidistant 
from  A  and  C.  Hence,  being  determined  by  these  points,  it  is  ±  to  AC, 
by  I,  prop.  XLI. 

2.    The  perpendicular  bisector  of  a  chord  passes  through  the 

center  of  the  circle  and  bisects  the  subtended  arcs. 

Eor  the  center  is  equidistant  from  the  ends  of  the  chord,  by  definition 
of  a  circle ;  .-.  it  lies  on  the  perpendicular  bisector  of  the  chord,  by 
I,  prop.  XLI. 


122  PLANE    GEOMETRY.  [Bk  III. 

Proposition  VI. 

185.  Theorem.  All  points  in  a  chord  lie  within  the  circle  ; 
and  all  points  in  the  same  line,  but  not  in  the  chord,  lie 
without  the  circle. 

0 


Given         the  points  Px  in  a  chord  AB,  and  P2  in  AB  pro- 
duced. 

To  prove    that  I\  is  within  the  circle,  and  P2  is  without. 

Proof.    1.   Suppose  0  the  center,  and  OA,  OB,  OP1}  OP2  drawn, 
and  031  A.  AB. 
Then  M  is  between  A  and  B.  Prop.  V 

2.  And  ' •  •  Z  A OM>  Z  Pl OM, 

.'.AO>P10,  I,  prop.  XX 

and  .-.  Px  is  within  the  O.  §  108,  def.  O,  cor.  3 

3.  And         v  Z  MOP,  >  Z  MOB, 

.-.  P20>  BO,  I,  prop.  XX 

and  .*.  P2  is  without  the  O.  §  108,  def.  O,  cor.  3 

Corollary.     A  straight  line  cannot  meet  a  circumference 

in  more  than  two  points. 

For  every  other  point  on  that  line  must  be  either  between  or  not 
between  those  two  points,  and  hence  must  lie  either  within  or  without 
the  circle. 


Exercises.     262.    Prove  that,  in  general,  two  chords  of  a  circle  can- 
not bisect  each  other.     What  is  the  exception  ? 

263.    What  is  the   locus  of  the  mid-points   of   a  pencil  of  parallel 
chords  of  a  circle  ?     Why  ? 


Prop.  VII. 


CHORDS   AND    TANGENTS. 


123 


Proposition  VII. 

186.  Theorem.  In  the  same  circle  or  in  equal  circles,  equal 
chords  are  equidistant  from  the  center;  and  of  tiro  unequal 
chords  the  greater  is  nearer  the  center. 


Given 


two  equal  ©  M.  M',  with  chords  AB  =  A'B",  AE  > 
A'B',  and  OC,  01),  O'C  _L's  from  center  0  to  AB, 
AE.  and  from  center  O1  to  A'B'. 


To  prove    that  (1)    OC=0'C,    (2)    OD  <  O'C 


Proof.    1. 


C,  C  bisect  AB,  A'B',  Prop.  V 

AC  =  A'C",  being  halves  of  equal  chords.       Ax.  7 
OA  =  O'A' 


Draw  OA,  O'A';  then 


and 


ZC  =  Z  C",  Prel.  prop.  I 

.'.  A  A  CO  ^  A  A' CO',    I.  prop.  XIX,  cor.  5 
and  OC  =  O'C,  which  proves  (1). 

3.  And  v  AE  >  A'B', 

then  AE  >  AB,  which  equals  A'B'.     Ax.  9 

4.  .'.  minor  AEE  >  AFB, 

so  that  E  does  not  lie  on  AEB.  Prop.  IV 

5.  And  V  0  and  AB  are  on  opposite  sides  of  AE, 

.-.  OC  cuts  AE,  as  at  G,  and  OD  <  OG. 

I,  prop.  XX 

6.  And       v  OG  <  OC,  r.  OD  <  OC.  Ax.  9 

7.  And       V  OC  =  O'C,  .-.  0/>  <  O'C".  Ax.  9 


124  PLANE   GEOMETRY.  [Bk.  III. 

Proposition  VIII. 

187.  Theorem.  In  the  same  circle  or  in  equal  circles, 
chords  that  are  equidistant  from  the  center  are  equal ;  and 
of  two  chords  unequally  distant,  the  one  nearer  the  center  is 
the  greater. 

Proof.  If  c,  c'  are  two  chords  of  the  same  circle  or  of  equal 
circles,  and  d,  d'  are  the  respective  perpendiculars  from 
the  center  upon  them ;  then  from  prop.  VII, 

If  c  >  c',     then     d  <  d', 

"    c  =  c',        "         d  =  d', 

"    c  <  c',        "         d  >  d'. 

Hence  the  converses  are  true  by  the  Law  of  Converse,  §  73. 

Corollary.     The  diameter  is  the  greatest  chord  in  a  circle. 
For  its  distance  from  the  center  is  zero. 


Exercises.  264.  AB  is  a  fixed  chord  of  a  circle,  and  XY  is  any 
other  chord  having  its  mid-point  P  on  AB.  What  is  the  greatest  and 
what  is  the  least,  length  that  XY  can  have  ? 

265.  What  is  the  locus  of  the  mid-points  of  equal  chords  of  a  circle  ? 

266.  Two  parallel  chords  of  a  circle  are  6  inches  and  8  inches,  respec- 
tively, and  the  distance  between  them  is  1  inch.     Find  the  radius. 

267.  Two  chords  are  drawn  through  a  point  on  a  circumference  so  as 
to  make  equal  angles  with  the  radius  drawn  to  that  point.  Prove  that 
the  chords  subtend  equal  arcs. 

268.  If  from  the  extremities  of  any  diameter  perpendiculars  to  any 
secant  are  drawn,  the  segments  between  the  feet  of  the  perpendiculars 
and  the  circumference  will  be  equal.     Draw  the  various  figures. 

269.  If  two  equal  chords  of  a  circle  intersect,  the  segments  of  the 
one  are  equal  respectively  to  the  segments  of  the  other. 

270.  Find  the  shortest  chord  which  can  be  drawn  through  a  given 
point  in  a  circle. 

271.  The  circumference  of  a  circle  whose  center  lies  on  the  bisector  of 
an  angle  cuts  equal  chords,  if  any.  from  the  arms. 


Prop.  IX.]  CHORDS   AND    TANGENTS.  125 

Proposition  IX. 

188.  Theorem.  Of  all  lines  passing  through  a  point  on  a 
circumference,  the  perpendicular  to  the  radius  drawn  to  that 
point  is  the  only  one  that  does  not  meet  the  circumference 
again. 


Given  point  P  on  the  circumference  of  a  O  with  center  0, 
and  AB,  PC,  respectively  perpendicular  and  oblique 
to  OP  at  P. 

To  prove  that  AB  does  not  meet  the  circumference  again,  but 
that  PC  does. 

Proof     1.  Let  031  A.  PC,  and  OX  be  any  oblique  to  AB. 

Then  031  <  OP,  I,  prop.  XX 

and  ..  31  is  within  the  O,  and  PC  cuts  the  circum- 
ference again.  §§  108,  109 

2.  Also,  OX  >  OP,  Why  ? 
and  .".  X,  any  point  except  P  on  AB.  is  without 
the  O.                                              §  108,  def.  O,  cor.  .3 

3.  .'.  the  perpendicular  does  not  meet  the  circumference 
again,  but  an  oblique  does. 

189.  Definitions.  The  unlimited  straight  line  which  meets 
the  circumference  of  a  circle  in  but  one  point  is  said  to  touch, 
or  be  tangent  to,  the  circle  at  that  point.  The  point  is  called 
the  point  of  contact,  or  point  of  tangency,  and  the  line  is  called 
a  tangent. 


126  PLANE    GEOMETRY.  [Bk.  III. 

A  tangent  from  a  point  to  a  circle  is  to  be  understood  as  the 
segment  of  the  tangent  between  the  point  and  the  circle. 

If  the  two  points  in  which  a  secant  cuts  a  circumference  continually 
approach,  the  secant  approaches  the  condition  of  tangency.  Hence  the 
tangent  is  sometimes  spoken  of  as  a  secant  in  its  limiting  position. 

Corollaries.  1.  One,  and  only  one,  tangent  can  be  drawn 
to  a  circle  at  a  given  point  on  the  circumference. 

For  the  tangent  is  perpendicular  to  the  radius  at  that  point,  and  there  # 
is  only  one  such  perpendicular.     (Has  this  been  proved  ?) 

2.  Any  tangent  is  perpendicular  to  the  radius  drawn  to  the 
point  of  contact.      (Why?) 

3.  A  line  perpendicular  to  a  radius  at  its  extremity  on  the 
circumference  is  tangent  to  the  circle.     (Why?) 

4.  The  center  of  a  circle  lies  on  the  perpendicular  to  any 
tangent  at  the  point  of  contact. 

For  the  radius  to  that  point  is  perpendicular  to  the  tangent,  and  as 
there  is  only  one  such  perpendicular  at  that  point  (prel.  prop.  II),  that 
perpendicular  must  be  the  radius. 

5.  The  perpendicular  from  the  center  to  a  tangent  meets  it 
at  the  point  of  contact. 

For  the  radius  to  that  point  is  perpendicular  to  the  tangent,  and  there 
is  only  one  perpendicular  from  the  center  to  the  tangent. 


Exercises.  272.  Show  that  of  these  three  properties  of  a  line,  (1)  the 
passing  through  the  center  of  a  circle,  (2)  the  being  perpendicular  to  a 
given  chord,  (3)  the  bisecting  of  that  chord,  any  two  in  general  necessi- 
tate the  third.     In  what  special  case  is  there  an  exception  ? 

273.  If  a  chord  is  bisected  by  a  second  chord,  and  the  second  by  a 
third,  and  the  third  by  a  fourth,  and  so  on,  the  points  of  bisection 
approach  nearer  and  nearer  the  center. 

274.  Tangents  drawn  to  a  circle  from  the  extremities  of  a  diameter 
are  parallel. 

275.  The  diameter  of  a  circle  bisects  all  chords  which  are  parallel  to 
the  tangent  at  either  extremity. 


Prop.  X.]  CHORDS  AND  TANGENTS.  127 

Proposition  X. 

190.  Theorem.  An  unlimited  straight  line  cuts  a  circum- 
ference, touches  the  circle,  or  does  not  meet  the  circle,  accord- 
ing as  its  distance  from  the  center  of  the  circle  is  less  than, 
equal  to,  or  greater  than,  the  radius. 


Given  OA,  OB,  OC,  the  perpendiculars  from  center  0  of 

O  M,  to  lines  S,  T,  X,   and  respectively  less  than, 
equal  to,  greater  than,  the  radius. 

To  prove    that  S  is  a  secant,  T  a  tangent,  N  a  line  not  meet- 
ing M. 

Proof.    1.  A,  B,  C  are  respectively  within  the  O,  on  the  circum- 
ference, or  without  the  O.  §  108,  def.  O,  cor.  3 

*2.  .'.  Sis  a  secant.  §  109,  2 

3.  And  T  is  a  tangent.  Prop.  IX,  cor.  3 

4.  N II  T.  I,  prop.  XVI,  cor.  3 

5.  And  .".  jV cannot  meet  O  M  because  it  cannot  cross  T. 

Corollary.      The  converses  are  true. 

Let  the  student  state  this  corollary  in  full,  and  show  that  the  Law  of 
Converse  (§  73)  applies. 

Exercises.  276.  What  is  the  locus  of  the  extremities  of  equal  tan- 
gents drawn  from  points  on  a  circumference  ? 

277.  Two  tangents  meet  at  a  point  the  length  of  a  diameter  distant 
from  the  center  of  the  circle.  How  many  degrees  in  their  included 
angle  ? 


* 


128  PLANE   GEOMETRY.  [Bk.  III. 


3.     ANGLES   FORMED  BY  CHORDS,   SECANTS, 
AND   TANGENTS. 

191.  Definitions.  A  segment  of  a  circle  is  either  of  the  two 
portions  into  which  the  circle  is  cut  by  a  chord. 

If  a  segment  is  not  a  semicircle,  it  is  called  a  major  or  a 
minor  segment  according  as  its  arc  is  a  major  or  minor  arc. 

E.g.  BBC  is  a  minor  segment,  and  BDE  is  a 
major  segment. 

The  fact  that  the  word  segment  is  used  to  mean 
a  part  of  a  line,  and  also  a  part  of  a  circle,  will 
not  present  any  difficulty,  since  the  latter  use  is 
rare,  and  the  sense  in  which  the  word  is  used  is 
always  evident.  It  means  "a  part  cut  off,"  and 
is  therefore  applicable  to  both  cases. 

192.  The  angle,  not  reflex,  formed  by  two  chords  which 
meet  on  the  circumference  is  called  an  inscribed  angle,  and  is 
said  to  stand  upon,  or  be  subtended  by,  the  arc  which  lies 
within  the  angle  and  is  cut  off  by  the  arms. 

It  is  also  called  an  angle  inscribed  in,  or  simply  an  angle  in,  the  segment 
whose  arc  is  the  conjugate  of  the  arc  on  which  it  stands. 

/.ABB  is  an  inscribed  angle,  standing  on  AB  ;  it  is  also  an  angle  in 
the  segment  BCBEA.  Similarly.  ABBA  is  in  the  segment  BAG  and 
stands  on  BA. 

193.  Points  lying  on  the  same  circumference  are  called 
coney  clic. 

Exercises.  278.  If  from  the  extremities  of  any  chord  perpendiculars 
to  that  chord  are  drawn,  they  will  cut  off  equal  segments  measured  from 
the  extremities  of  any  diameter.  (Draw  a  perpendicular  from  the  center 
to  the  chord.) 

279.  If  a  tangent  from  a  point  B  on  a  circumference  meets  two  tan- 
gents from  A,  C,  on  the  circumference,  in  points  X,  Y ;  and  if  the  lines 
joining  the  center  to  A,  X,  Y,  C,  are  a,  x,  ?/,  c,  respectively,  then  Zxy  = 
/ax  -\-  /  ye,  and  XY  =  AX  +  YC. 


Prop.  XI.] 


CHORDS  AND    TANGENTS. 


129 


Proposition  XI. 

194.    Theorem.     An  inscribed  angle  equals  half  the  central 
an; jle  standing  on  the  same  arc. 


Fig.  2. 


Fio.  3. 


Given 


AVB  an  inscribed  angle,  and  AOB  the  central  angle 
on  the  same  arc  AB. 


To  prove    that  Z  AVB  =  £  Z  A  OB. 

Proof.    1.  Suppose  VO  drawn  through  center  0,  and  produced 
to  meet  the  circumference  at  X. 
Then        Z  XVB  =  Z  VBO.  I,  prop.  Ill 

2.  And  Z  XOB  =  Z  XVB  +  Z  FJ50,  Why  ? 

=  2  Z  XTO.  Step  1 

3.  r.A.XVB  =  \£XOB.  Ax.  7 

4.  Similarly  Z .!  FA  =  JZ  A  OX  (each  =  zero  in  Fig.  2), 
and       .•.ZJF5  =  iZJ05.  Ax.  2 

The  proof  holds  for  all  three  figures,  point  A  having  moved 
to  X  (Pig.  2),  and  then  through  X  (Pig.  3). 

195.  The  theorem  is  often  stated  thus :  An  inscribed  angle 
is  measured  by  half  its  intercepted  arc 

This  expression,  like  that  mentioned  in  §  180  is  not  strictly  correct. 
The  angle  and  the  arc  simply  have  the  same  numerical  measure  as  proved 
later  in  §  254. 


130  PLANE    GEOMETRY.  [Bk.  IT1. 

Corollaries.  1.  Angles  in  the  same  segment,  or  in  equal 
segments,  of  a  circle  are  equal.      (Why  ?) 

2.  If  from  a  point  on  the  same  side  of  a  chord  as  a  given 
segment,  lines  are  drawn  to  the  ends  of  that  chord,  the  angle 
included  by  those  lines  is  greater  than,  equal  to,  or  less  than, 
an  angle  in  that  segment,  according  as  the  point  is  within,  on 
the  arc  of,  or  without,  the  segment. 

This  follows  from  cor.  1  and  from  I,  prop.  IX.  Draw  the  figure  and 
prove. 

3.  The  converse  of  cor.  2  is  true  by  the  Law  of  Converse. 
Hence  the  locus  of  the  vertex  of  a  constant  angle  whose  arms 
pass  through  two  fixed  points  is  an  arc. 

Let  the  student  state  the  converse  in  full,  and  give  the  proof. 


Exercises.     280.     In  the  figures  on  p.  129,  prove  that  if  P  is  taken 
'    anywhere  on  BY,  then  Z  PBV  +  Z  BVP  is  constant. 

281.  In  Fig.  3,  p.  129,  if  BO  is  produced  to  meet  the  circumference 
at  W,  and  the  point  of  intersection  of  BW  and  AV  is  called  F,  prove 
that  A  YVB  and  "MM Fare  mutually  equiangular. 

282.  What  is  the  locus  of  the  vertex  of  a  triangle  on  a  given  base 
and  with  a  given  vertical  angle  ?     Prove  it. 

283.  In  Fig.  1,  p.  129,  suppose  A  to  move  freely  on  the  arc  VAXB, 
and  suppose  A  AVB,  VBA  bisected  by  lines  meeting  at  P.  Show  that 
the  locus  of  P  is  a  constant  arc. 

284.  If  the  vertices  of  a  hexagon  are  concyclic,  the  sum  of  any  three 
alternate  interior  angles  is  a  perigon.  (That  is,  the  sum  of  three  angles, 
taking  every  other  one.) 

285.  Two  equal  chords  with  a  common  extremity  are  symmetric  with 
respect  to  the  diameter  through  that  extremity,  as  an  axis  ;  so  also  are 
their  corresponding  arcs. 

286.  If  from  any  point  P.  on  the  diameter  AB,  PX  and  PY  are 
drawn  to  the  circumference  on  the  same  side  of  AB  and  making 
ZXP^l  =  ZBPY.  then  &  APX  and  YPB  are  mutually  equiangular. 

287.  If  any  number  of  triangles  on  the  same  base  and  on  the  same  side 
of  it  have  equal  vertical  angles,  the  bisectors  of  the  angles  are  concurrent. 

288.  Prove  that  two  chords  perpendicular  to  a  third  chord  at  its 
extremities  are  equal. 


Prop.  XII.] 


CHORDS   AND    TANGENTS. 


131 


Proposition  XII. 

196.  Theorem.  An  angle  in  a  segment  is  greater  than, 
equal  to,  or  less  than,  a  right  angle,  according  as  the  segment 
is  less  than,  equal  to,  or  greater  than,  a  semicircle. 


Given 


the  segments  ABE,  ACE,  ABE  of  a  circle  with 
center  0,  respectively  less  than,  equal  to.  greater 
than,  a  semicircle. 


To  prove    that  A  AEB,  AEC,  AEB  are  respectively  greater 
than,  equal  to,  less  than,  a  right  angle. 

Proof.    1.  Draw  OB,  OB. 

Then  v  Z  AEB  =  \  reflex  AAOB,  Prop.  XI 

.'./.AEB  >  rt.  Z. 
'  Z  AEC  =  i  st.  Z  AOC,  Why  ? 

..ZAEC=  it.  Z. 
Z  A  EB  =  \  oblique  ZAOB,  Why  ? 

.-.A  AEB  <  rt,  Z. 


2.  And 


And 


Corollary.  A  segment  is  less  than,  equal  to,  or  greater 
than,  a  semicircle,  according  as  the  angle  in  it  is  greater  than, 
equal  to,  or  less  than,  a  right  angle. 

From  prop.  XII,  by  the  Law  of  Converse,  §  73.  Let  the  student  write 
out  the  proof. 

Note.     The  discovery  that  an  angle  in  a  semicircle  is  a  right  angle  is 

attributed  to  Thales,  who,  tradition  asserts,  sacrificed  an  ox  to  the  gods 
in  honor  of  the  event. 


132 


PLANE    GEOMETRY. 


[Bk.  III. 


Propositiox  XIII: 

197.  Theorem.  An  angle  formed  by  a  tangent  and  a  chord 
of  a  circle  equals  half  of  the  central  angle  standing  on  the 
intercepted  arc. 

C 


Given 


AB  a  chord,  XX'  a  tangent  through  A,  and  0  the 
center  of  the  circle. 


To  prove    that  Z  XAB  =  i  Z  A  OB, 

and         ZBAX'  =  iZBOA. 

Proof.    1.  Produce  AO  to  meet  the  circumference  at  C. 
Then       A  XAC  =\AAOC, 

=  \  st.  Z.  Why  ? 

2.  And    •.'  ZBAC  =±ZBOC,  Why? 

.-.  Z  XAB  =  i  Z  AOB.  Ax.  3 

3.  Also,  V  Z  C^X'  =  |  Z  COJ,  Why  ? 

.-.  Z  #4X'  =  }Z  1*0.4.  Ax.  2 

Corollary.  Tangents  to  a  circle  from  the  same  external 
jjoint  are  equal. 

For,  connect  the  points  of  tangency,  and  two  angles  of  the  triangle 
are  equal  by  this  theorem. 

198.  The  theorem  is  often  stated  thus:  An  angle  formed  by 
a  tangent  and  a  chord  of  a  circle  is  measured  by  half  its  inter- 
cepted arc. 

See  §  195. 


&Y 


Prop.  XIV.] 


CHORDS  AND    TANGENTS. 


L33 


Proposition  XIV. 

199.  Theorem.  An  angle  formed  by  two  unlimited  inter- 
secting lines  which  meet  the  circumference  equals  either  the 
sum  or  the  difference  of  half  the  central  angles  on  the  inter- 
cepted arcs,  according  as  the  point  of  intersection  is  within  or 
without  the  circle. 


Fig.  1. 


Fig.  2. 


Given  two  lines  XX',  YY'  meeting  a  circumference  at 
A,  A'  and  B,  B',  respectively,  and  intersecting  at  P. 

To  prove  that  Z  A'PB'  equals  half  the  central  angle  on  A'B' 
plus  or  minus  half  that  on  AB,  according  as  P  is 
within  or  without  the  circle. 

Proof.         Suppose  AB'  drawn. 

Then  Z  A'PB'  =  Z  A'AB'  ±  Z  AB'B.      §  88 

==  J  cent,  Z  on  AB'  ±  |  cent.  Z  on  AB. 

Prop.  XI 

The  theorem  is  thus  re-stated  for  two  of  the  special  cases : 

Corollaries.     1.  An  angle  formed  by  two  chords  equals  the 
sum,  of  half  the  central  angles  on  the  intercepted  arcs. 
See  Fig.  1.     (State  this  as  suggested  in  §  195.) 

2.  An  angle  formed  by  two  secants  intersecting  without  tin- 
circle  equals  the  difference  of  half  the  central  angles  on  the 
intercepted  arcs. 

See  Fig.  2.     (State  this  as  suggested  in  §  195.) 


134 


PLANE    GEOMETRY. 


[Bk.  III. 


Prop.  XIV  is,  of  course,  true  for  tangents  as  well  as  chords 
and  secants.     The  following  figures  represent  special  cases. 


Fig.  3. 


Fig.  4. 


Fig.  5. 


Fig.  6. 


Fig.  7. 

Fig.  3  is  a  special  case  where  P  is  at  0,  and  merely  affirms  that 
a  central  angle  equals  itself.  Fig.  4  shows  that  prop.  XI  is  a  special  case 
of  prop.  XIV.     Fig.  6  shows  the  same  for  prop.  XIII. 

Corollary.  3.  An  angle  formed  by  a  secant  and  a  tan- 
gent, or  by  two  tangents,  equals  the  difference  of  half  the 
central  angles  on  the  intercepted  arcs. 

See  Figs.  7  and  8. 


Prop.  XV.] 


CHORDS  AND    TANGENTS. 


135 


Proposition  XV. 

200.   Theorem.     If  two  parallel  liar*  intercept  arcs  on   a 
circumference,  those  arcs  are  equal. 


Given         two  parallel  lines.  I  and  II.  intercepting  arcs  AB. 
' '  I>.  on  the  circumference  of  a  circle  with  center  0. 

To  prove    that  AB  =  CI). 

Proof.    1.  Suppose  YOY  _L  I.  and  to  cut  BC  at  31  Fig.  1. 

Then  YY'  ±  II.  I.  prop.  XVII,  cor-  1 

2.  And         BM  =  MC,  and  AM=  JIB.  Prop.  V 

3.  .'.AB=CD.  Ax.  3 

Note.  The  proof  is  the  same  for  Figs.  2.  3  ;  in  Fig.  2.  BC  equals  zero; 
and  in  Fig.  3,  DA  also  equals  zero.  It  should  be  noticed  that  Figs.  1,  2, 
3,  respectively,  may  be  considered  as  special,  or  at  least  as  limiting  cases 
of  Figs.  2.  7.  and  8  of  prop.  XIV.  In  prop.  XIV  as  P  moves  farther 
and  farther  to  the  right  the  lines  come  nearer  and  nearer  to  being  parallel, 
the  angle  A PB  approaches  nearer  and  nearer  zero,  and  hence  the  cen- 
tral angles  on  arcs  BA,  A'B'  approach  nearer  and  nearer  equality.  It 
might  therefore  be  inferred  that  when  the  lines  become  parallel,  the  arcs 
become  equal,  as  proved  in  prop.  XV. 


Exercise.  289.  The  chords  which  join  the  extremities  of  two  equal 
arcs  are  either  parallel,  or  else  they  intersect  and  are  equal  and  cut  off 
equal  segments  from  each  other. 


136 


PLANE    GEOMETRY. 


[Bk.  III. 


INSCRIBED   AND   CIRCUMSCRIBED   TRIANGLES    AND 
QUADRILATERALS. 


201.  Definitions.  If  the  ver- 
tices of  the  angles  of  a  poly- 
gon lie  on  a  circumference, 
the  polygon  is  said  to  be  in- 
scribed in  the  circle,  and  the 
circle  is  called  a  circumscribed 
circle. 


Inscribed  quadrilateral. 
Circumscribed  circle. 


If  the  lines  of  the  sides  of 
a  polygon  are  tangent  to  a 
circle,  the  polygon  is  said  to 
be  circumscribed  about  the  cir- 
cle, and  the  circle  is  called 
an  inscribed  or  escribed  circle, 
according  as  it  lies  within  or 
without  the  polygon. 


Circumscribed  quadrilateral. 
Inscribed  circle. 


Inscribed  cross  quadrilateral. 
Circumscribed  circle. 


Circumscribed  quadrilateral. 
Escribed  circle. 


The  words  inscriptible,  circumscriptible,  escriptible  mean  capable  of 
being  inscribed  in,  circumscribed  about,  escribed  to,  a  circle. 


Exercise.  290.  Tf  any  two  chords  cut  within  the  circle,  at  right 
angles,  the  sum  of  the  squares  on  their  segments  equals  the  square  on 
the  diameter. 


Prop.  XVI.]  VIBCLE8  AND  POLYGONS.  L37 

Proposition  XVI. 

202.  Theorem.  A  circumference  can  be  described  to  pass 
through  the  three  vertices  of  any  triangle.  (Circumscribed 
circle.) 

c 


Given         the  points  A,  B,  C,  the  vertices  of  A  ABC. 

To  prove    that  a  circumference  can  be  described  to  pass  through 
A,  B,  C. 

Proof.    1.   There  is  such  a  circumference.  $  131,  cor.  2 

2.  And  the  center  of  the  O  can  be  found.    §  131,  cor.  1 

Note.     The  relation  between  prop.  XVI  and  prop.  XVII  should  be 
noticed.     Similarly  for  props.  XVIII  and  XIX,  and  for  XX  and  XXI. 


Exercises,  291.  Prove  from  prop.  XVI  and  prop.  XI  that  the  sum  of 
the  interior  angles  of  any  triangle  equals  a  straight  angle. 

292.  If  the  hypotenuse  of  a  right-angled  triangle  is  the  diameter  of  a 
circle,  the  circumference  passes  through  the  vertex  of  the  right  angle. 
(Corollary.  The  median  from  the  vertex  of  the  right  angle  of  a  right- 
angled  triangle  equals  half  of  the  hypotenuse.) 

293.  A  line-segment  of  constant  length  slides  so  as  to  have  its  extremi- 
ties constantly  resting  on  two  lines  perpendicular  to  each  other.  Find 
the  locus  of  its  mid-point. 

294.  If  a  circle  is  described  on  the  line  joining  the  orthocenter  to  any 
vertex,  as  a  diameter,  prove  that  the  circumference  passes  through  the 
feet  of  the  perpendiculars  from  the  other  vertices  to  the  opposite  sides. 

295.  Prove  that  the  perpendiculars  from  the  vertices  of  a  triangle  to 
the  opposite  sides  bisect  the  angles  of  the  triangle  formed  by  joining  their 
feet ;  the  so-called  Pedal  Triangle. 


138 


PLANE    GEOMETRY. 


[Bk  III. 


Proposition  XVII. 

203.    Theorem.     A  circle  can  be  described  tangent  to  the 
three  lines  of  any  triangle.     {Inscribed  and  escribed  circles.) 


t 


Given         the  lines  a,  b,  c,  forming  a  A  ABC. 

To  prove    that  a  circle  can  be  described  tangent  to  a,  b,  c. 

Proof.    1.  Let  0  be  the  in-center,  Ox,  02,  03  the  ex-centers. 
Let  OP,  OQ,  OB±  a,  b,  c. 

Then  A  AB  0  5£  A  A  Q  0, 

and  ABBO^ABPO.      I,  prop.  XIX,  cor.  7 

2.  .-.  OQ  =  OB=  OP.  Why  ? 

3.  .'.  P,  Q,  B  are  concyclic  §  108,  def.  O,  cor.  3 

4.  And  V  AB  _L  OB,  AB  is  a  tangent. 

Prop.  IX,  cor.  3 
Similarly,  a,  b,  c,  are  tangent  to  the  other  three  (D. 

Corollary.     A    circle   can   be    described   tangent   to  three 
lines  not  all  parallel  nor  concurrent. 


Prop.  XVIII.] 


CIRCLES   AND   POLYGONS. 


139 


Proposition    XVIII. 

204.  Theorem.  In  an  inscribed  quadrilateral  the  swn  or 
difference  of  two  opposite  angles  equals  the  sum  or  differ- 
ence of  the  other  two  opposite  angles,  according  as  the  quad- 
rilateral is  convex  or  cross. 


Given         the  inscribed  convex  quadrilateral  ABCD. 

To  prove    that  in  Fig.  l,Zi  +  ZC  =  Z5  +  ZD. 

Proof  for  Fig.  i.    1.  Z  A  +  Z  C  =  \  central  Z  on  1W  +  DB 

=  st.  Z.  Prop.  XI 

2.  Similarly,  Z  B  +  Z  i>  =  st.  Z. 

3.  ,'.Zi  +  Z(J  =  Z5  +  ZJ).  §  30 

Proof  for  Fig.  2.    If   the    quadrilateral    is    cross,    Z.  C  —  Z.  A 
=  Z.D  —  /LB,  since  each  equals  zero.     Why  ? 

Corollaries.  1.  A  parallelogram  inscribed  in  a  circle  has 
all  of  its  angles  equal,  and  is  therefore  a  rectangle.      (Why  ?) 

2.  The  opposite  angles  of  an  inscribed  convex  quadrilateral 
are  supplemental. 


Exercises.  296.  In  the  figure  of  prop.  XIII,  if  P  is  the  mid-point 
of  arc  AB,  prove  that  P  is  equidistant  from  AX  and  AB.  Suppose  the 
arc  BCA  is  taken,  instead  of  AB. 

297.  If  a  circle  is  described  on  one  side  of  a  triangle  as  a  diameter, 
prove  that  the  circumference  passes  through  the  feet  of  the  perpendiculars 
drawn  to  the  other  two  sides  from  the  opposite  vertices. 


140 


PLANE    GEOMETRY 


[Bk.  III. 


Proposition   XIX. 

205.  Theorem.  In  a  circumscribed  quadrilateral  the  sum 
or  difference  of  two  opposite  sides  equals  the  sum  or  differ- 
ence of  the  other  two  opposite  sides,  according  as  the  quadri- 
lateral is  convex  or  cross. 

b, 


Fig.  1.  Fig.  2. 

Given         the  circumscribed  convex  quadrilateral  abed. 
To  prove    that  in  Fig.  1,  a  +  c  =  b  +  d. 
Proof  for  Fig.  z,  as  lettered. 

1.  ax  =  d2,    a2  =  bx,  cx  =  b2,  c2  =  dx.      Prop.  XIII,  cor. 

2.  .'.  at  +  a2  +  cx  +  c2  —  bx  +  b2  +  r/x  +  d2.  Ax.  2 
.3.  Or,                   a'-\-c  =  b  +  d.  Ax.  8 

Proof  for  Fig.  2.   If  the  quadrilateral  is  cross,  c  —  a  =  d  —  b. 

1.  v  cx  =  b2,  and  c2  =  dx,  .'.  c  =  />.:  +  r^. 

2.  v  ax  =  r/2,  and  a2  =  bx,  .'.  a  =  bx  +  d2. 

3.  .' .  r  —  a  =  d  —  b. 


Ax.  3 


Corollary.     ^4  parallelogram  circumscribed  about  a  circle 
has  all  of  its  sides  equal,  and  is  therefore  a  rhombus.    (Why?) 


Exercises.  298.  The  bisector  of  an  angle  formed  by  a  tangent  and 
chord  bisects  the  intercepted  arc. 

299.  Given  two  pairs  of  parallel  chords,  ABWA'IV,  and  BCW  B'C ; 
prove  that  AC'  WA'C. 


Prop.  XX.J  CIRCLES   AND   POLYGONS.  141 

Proposition  XX. 

206.  Theorem.  If  the  sum  of  two  opposite  angles  of  a 
quadrilateral  equals  the  sum  of  the  other  two  opposite  angles, 
the  quadrilateral  is  inscriptible. 


Given         the  quadrilateral  ABCD  such  that 
Z.A  +  Z.C  =  £B  +  £D. 
To  prove    that  ABCD  is  iuscriptible. 

Proof.  1.  Suppose  the  circumference  determined  by  A,  B,  C  not 
to  pass  through  D,  but  to  cut  CD  at  E.  Prop.  XVI 
Draw  AE.     Then  £B  +  £  AEC  =  ZC  +  Z  BAE. 

Prop.  XVI I  [ 
2.  But  Z  B  +  Z2>  =  Z  C  +  A  A, 

and  .-.  Z  AEC  -  Z  D  =  Z  ^J^  -  Z  /I, 
or,  Z  ^JZ>  =  -  Z  £\m       I,  prop.  XIX 

But  this  is  absurd  :  hence  step  1  is  absurd. 
The  proof  is  the  same  for  D'. 

Corollary.      If  two  opposite  angles  of  a  quadrilateral  are 
supplemental,  the  quadrilateral  is  inscriptible. 


Exercises,     300.    A  square  is  inscriptible. 

301.  Every  equiangular  quadrilateral  is  inscriptible. 

302.  The  intersection  of  the  diagonals  of  an  equiangular  quadrilateral 
is  the  center  of  the  circumscribed  circle. 

303.  A  circle  is  described  on  one  of  the  equal  sides  of  an  isosceles 
triangle  as  a  diameter.     Prove  that  the  circumference  bisects  the  base. 


142  PLANE    GEOMETRY.  [Bk.  III. 

Proposition  XXI. 

207.  Theorem.  If  the  sum  of  two  opposite  sides  of  a 
quadrilateral  equals  the  sum  of  the  other  two  opposite  sides, 
the  quadrilateral  is  circumscriptible. 

D   E      p/ 


A 


"B 


Given         the  quadrilateral  ABCD  such  that 

AB  +  CD  =  BC  +  DA. 
To  prove    that  ABCD  is  circumscriptible. 

Proof.    1.  Suppose  the  O  tangent  to  AB,  BC,  CD  not  to  be 
tangent  to  DA,  but  to  be  tangent  to  EA. 

Prop.  XVII 
Then  AB  +  CE  =  BC  +  EA.  Prop.  XIX 

2.  But  AB  +  CD  =  BC  +  DA,  Given 

and  .'.  CD  -  CE,  or  ED,  =  DA-  EA.  Ax.  3 

But  this  is  absurd ;  hence  step  1  is  absurd. 

I,  prop.  VIII,  cor. 
The  proof  is  the  same  for  D'. 


Exercises.     304.    A  square  is  circumscriptible.      (Notice  the  relation 
between  exs.  300-302  and  exs.  304-306.) 

305.  Every  equilateral  quadrilateral  is  circumscriptible. 

306.  The  intersection  of  the  diagonals  of  an  equilateral  quadrilateral 
is  the  center  of  the  inscribed  circle. 

307.  A',  B"  are  the  feet  of  perpendiculars  from  A .  B  on  a.  b  in  A  ABC  ; 
M  is  the  mid-point  of  AB.     Prove  that  Z  B'A'M  -  Z  MB' A'  =  AC. 


Sec.  208.T  TW0    CIRCLES.  14^ 


5.     TWO   CIRCLES. 

208.  Definitions.  Two  circles  are  said  to  touch  or  to  be  tan- 
gent when  their  circumferences  have  one,  and  only  one,  point 
in  common. 

They  are  said  to  be  internally  or  externally  tangent  according  as  one 
circle  lies  within  or  without  the  other.  The  more  accurate  expression, 
a  tangent  circumference,  is  often  used  instead  of  a  tangent  circle. 

The  line  determined  by  the  centers  of  two  circles  is  called 
their  center-line ;  the  segment  of  the  center-line,  between  the 
centers,  is  called  their  center-segment. 

If  two  circles  have  a  common  center,  they  are  said  to  be 
concentric. 

The  expression  concentric  circumferences  is  also  used. 


Exercises.  308.  A  triangle  is  inscribed  in  a  circle.  Prove  that  the 
sum  of  three  angles,  one  in  each  segment  of  the  circle,  exterior  to  the 
triangle,  equals  a  perigon. 

309.  Prove  that  a  perpendicular  from  the  orthocenter  of  a  triangle 
to  a  side,  produced  to  the  circumference  of  the  circumscribed  circle,  is 
bisected  by  that  side. 

310.  Prove  that  the  bisectors  of  any  angle  of  an  inscribed  quadri- 
lateral and  the  opposite  exterior  angle  meet  on  the  circumference. 

311.  If  the  diagonals  of  an  inscribed  quadrilateral  bisect  each  other, 
what  kind  of  a  quadrilateral  is  it  ? 

312.  Prove  that  if  two  consecutive  sides  of  a  convex  hexagon  inscribed 
in  a  circle  are  respectively  parallel  to  their  opposite  sides,  the  remaining 
sides  are  parallel  to  each  other. 

313.  Prove  that  the  bisectors  of  the  angles  formed  by  producing  the 
opposite  sides  of  an  inscribed  quadrilateral  to  meet,  are  perpendicular  to 
each  other.     (A  proof  may  be  based  on  cors.  1  and  2  of  prop.  XIV.) 

314.  Prove  that  if  the  diagonals  of  an  inscribed  quadrilateral  are 
perpendicular  to  each  other,  the  line  through  their  intersection  perpen- 
dicular to  any  side  bisects  the  opposite  side.     (Brahmagupta's  theorem.) 


144 


PLANE    GEOMETRY. 


[Bk.  III. 


Proposition  XXII. 

209.  Theorem.  If  two  circumferences  meet  in  a  'point 
which  is  not  on  their  center-line,  then  (1)  they  meet  in  one 
other  point,  (2)  their  center-line  is  the  perpendicular  bisector 
of  their  common  chord,  (3)  their  center-segment  is  greater 
than  the  difference  and  less  than  the  sum  of  the  radii. 


Fig.  1.  Fig.  2. 

Given         31  and  iV,  two   circumferences  with  centers  A,  B, 
meeting  at  P  not  on  AB. 

To  prove    that  (1)  they  meet  again,  as  at  P'; 

(2)  AB  _L  PP'  and  bisects  it,  as  at  C\ 

(3)  AB  >  the  difference  between  AP  and  BP 

and  <  AP  +  BP. 

Proof.    1.  In  Fig.  1,  suppose  AABP  revolved  about  AB  as  an 
axis  of  symmetry,  thus  determining  A  AP'B. 
Then  V  AP1  =  AP,  and  BP'  =  BP, 

.'.  P'  is  on  both  M  and  N,  which  proves  (1). 

§  108,  def.  O,  cor.  3 

2.  In  Fig.  2,  V  AP  =  AP,  and  BP  =  BP,       §  109,  1 

3.  .'.  A  and  B  lie  on   the  _L  bisector  of  PP',  which 
proves  (2).  I,  prop.  XLI 

4.  AB  >  the   difference    between    AP   and    BP   and 
<  AP  +  BP,  which   proves   (3).  §  75  and  cor. 

Corollary.     If  two  circumferences  meet  at  one  point  only^ 
that  point  is  on  their  center-line.     \  Why  ?) 


Prop.  XXIIL]  TWO    CIRCLE*.  145 

Proposition  XXIII. 

210.   Theorem.     If   two  circles    meet  on    their   center-line, 
they  are  tangent. 

P 


0       0'  A 

Given  0  and  O',  the  centers  of  two  circles  with  radii  OA, 

0'A}  which  meet  on  their  center-line  at  A. 

To  prove     that  the  circles  are  tangent. 

Proof.    1.  Let  P  be  any  point,  other  than  A,  on  circumference 
with  center  0,  and  draw  OP,  OP. 
Then       00'  +  O'P  >  OP  or  its  equal  OA. 

I,  prop.  VIII 

2.  And  v  00'=  OA  -  O'A, 

.-.  OA-  0'A+  OP>  OA, 
or  OP  >  OA, 

by  adding  O'A  and  subtracting  OA.  Axs.  4.  5 

3.  .".  P  is  without  the  circle  with  center  0'. 

§  108.  def.  O,  cor.  3 

4.  And  ".'  the    ©    have    only    one   point    in    common, 
.'.  they  are  tangent.  §  208 

Corollaries.     1.    If  two  circumferences   intersect,  neither 
'point  of  intersection  is  on  the  center-line.      (Why?) 

2.  If  two  circles  touch,  they  have  a  common  tangent-line  at 
the 'point  of  contact. 

For  a  perpendicular  to  their  center-line  at  that  point  is  tangent  to 
both.     (Why  ?) 

Exercise.     315.    Find  the  locus  of  the  centers  of  all  circles  tangent 
to  a  given  circle  at  a  given  point. 


y. 


146 


PLANE    GEOMETRY. 


[Bk.  111. 


6.     PROBLEMS. 
Proposition  XXIV. 
211.    Problem.      To  bisect  a  given  arc. 


Solution.    1.  Draw  its  chord  AB. 

2.  Draw  PCI  AB  at  its  rnid-point. 
Then  PC  bisects  the  arc. 


§  28 

§§  114,  116 

Prop.  V,  cor.  2 


Proposition  XXV. 

212.    Problem.      To  find   the   center   of  a   circle,  given  its 
circumference  or  any  arc. 


Given         a  circumference,  or  an  arc  ABC. 

Required    to  find  the  center  of  the  circle. 

Solution.    1.  Draw  two  chords  from  B,  as  BA,  BC.  §  28 

2.  Draw  their  _L  bisectors  DD\  EE\  §§  114,  116 

intersecting  at  the  center  O.  §  131,  cors.  1,  4 

Note.     Hereafter  it  will  be  assumed  that  the  center  is  known  if  an 
arc  is  known,  for  it  may  always  be  found  by  this  problem. 


Prop.  XXVI]  TWO   CIRCLES.  147 

Proposition  XXVI. 

213.    Problem.      To  draw  a  tangent  to  a  given  circle  from 
a  given  point. 

1.  If  the  point  is  on  the  circumference. 

Solution.    1.  At  the  given  point  erect  a  perpendicular  to  the 
radius  drawn  to  the  point.  I,  prop.  XXIX 

2.   This   is  the  required   tangent,  and  the   solution  is 
unique.  Prop.  IX,  cors.  3,  1 

2.  If  t tie  point  is  without  tlie  circle. 


Given         a  circle   PP'B,   with    center    0;    also    an  external 
point  A. 

Required    from  A  to  draw  a  tangent  to  O  PPB. 

Construction.    1.  Draw  AO.  §28 

2.  Bisect  AO  at  M.  I,  prop.  XXXI 

3.  Describe  a  O  with  center  3L  radius  MO.  §  109 

4.  Join  A  to  intersections  of  circumferences.  §  28 
Then  these  lines,  AP,  AP',  are  the  required  tangents. 

Proof.    1.  The  circumferences  will  have  two  points  in  common, 
and  only  two.      Prop.  XXII;  I.  prop.  XLIIL  cor.  3 

2.  And  v  A  APO,  OP' A  are  rt.  A,  Why  ? 

.'.  AP,  AP'  are  tangents.  Why  ? 

(Would  this  solution  hold  for  case  1  ?) 


148 


PLANE    GEOMETRY. 


[Bk.  III. 


Proposition  XXVII. 

214.   Problem.     To  draw  a  common  tangent  to  two  given 
circles. 


Fig.  1, 


Fig.  2. 


Given         two  circles  A,  B,  with  radii  r,  r'  (r  >  ?•'),  and  centers 
0,  0',  respectively. 

Required    to  draw  a  common  tangent  to  them. 

Construction.     1.  Describe  ©  Ax,  A%  (Figs.  1  and  2),  with  centers 
0,  and  radii  /•  —  r'  and  r  +  >•',  respectively.        §  109 

2.  From  0'  draw  tangents  0'C\,  0'CSJ  to  ©  Ax,  A*. 

Prop.  XXVI 

3.  Draw  OC\,  OC2,  cutting  circumferences  A  at  Ex,  E2. 

§  28 

4.  Draw  0'DX  II  OEx,  and  O'D,  II  ^aO.  §  IIS 

5.  Draw  ^7)^  E2D2;  they  are  the  tangents. 

Proof.    1.  Zs  C1?  C2  are  rt.  A  Why? 

2.  In  Fig.  1, 

V  CXEX  =  0^  -  0CX  =  /•  -  (;•  -  r')  =  r', 

.'.  CXEX  =  and  II  0'DX.  Const.  1,  4 

3.  .'.  CxO'DxEx  is  a  □,  and  A  Ex,  1\  are  rt.  A, 

I,  props.  XXV.  XXIII.  cor. 
and  .'.  DXEX  is  tangenl  to  CD  A.  /»'.       Prop.  IX.  cor.  3 


Prop.  XXVII.] 


TWO   CIRCLES. 


149 


Similarly,  in  Fig.  2,  E2C2  =  and  II  D20',  and  E2D20'C2  is  a 
□  ,  and  D2E2  is  a  tangent.  In  both  figures  a  second  tangent 
can  evidently  be  drawn,  the  solution  being  analogous  to  that 
above  given.    Hence  there  are  four  tangents  in  general. 

Note.     The  following  special  cases  are  of  interest. 


Fig.  3. 


Fig.  4. 


(3 


Fig.  5. 


Fig. 


In  Fig.  3  the  two  circles  have  moved  to  external  tangency,  and  the 
two  interior  tangents  have  closed  up  into  one.  In  Fig.  4  the  circumfer- 
ences intersect  and  the  interior  tangents  have  vanished.  In  Fig.  5  the 
circles  have  become  internally  tangent  and  the  two  exterior  tangents  have 
closed  up  into  one.  In  Fig.  6  the  circle  B  lies  wholly  within  the  circle  A, 
and  the  tangents  have  all  vanished.  In  all  cases  the  center-line  is  evi- 
dently an  axis  of  symmetry. 


Exercises.  316.  All  tangents  drawn  from  points  on  the  outer  of  two 
concentric  circumferences  to  the  inner  are  equal. 

317.  Find  the  locus  of  the  centers  of  all  circles  touching  two  intersect- 
ing lines.  (Show  that  it  is  a  pair  of  perpendiculars.)  Suppose  the  two 
lines  were  parallel  instead  of  intersecting. 


150  PLANE    GEOMETRY.  [Bk.  III. 

Proposition   XXVIII. 

215.    Problem.      On  a  given  line-segment  as  a  chord  to  con- 
struct a  segment  of  a  circle  containing  a  given  angle. 


c 

0 

)C 

m 

^y 

\ 

0' 

r 

Given         the  line-segment  AB  and  the  A  N. 

Required    on  AB  to  construct  a  segment  of  a  circle,  containing 
AN. 

Construction.    1.  Draw  BD   and  AC,   making  A  ABD,  BAC 
equal  to  A  N.  I,  prop.  XXXII 

2.  Draw  YY',  the  X  bisector  of  AB.       I,  prop.  XXXI 

3.  Draw  J_'s  to  ^C,  BD,  from  J,  #.        I,  prop.  XXIX 

These  _L's  will  intersect  YY'  at  the  centers  of  the 
©  whose  segments  on  AB  are  required. 

Proof.    1.  The  two  J_'s  from  A,  B,  meet  YY',  as  at  0',  0. 

I,  prop.  XVII,  cor.  4 

2.  0  is  the  center  of  O  with  chord  AB  and  tangent  BD. 

Prop.  V,  cor.  2 ;  prop.  IX,  cor.  4 

3.  .'.A  ABD,  or  Z  N,  =  £  central  Z  on  AEB, 

Prop.  XIII 
=  Z  in  segment  AB  Y. 

Prop.  XI 
Similarly  for  segment  Y'BA,  where  ABAC  =  £ 
central  Z  on  the  intercepted  arc. 


Sec  216. 


TWO    CIRCLES.  151 


216.  Definitions.  Two  intersecting  arcs  are  said  to  form  an 
angle,  meaning  thereby  the  angle  included  by  their  respective 
tangents  at  the  point  of  intersection. 

An  arc  and  a  secant  are  said  to  form  an  angle,  meaning  thereby 
the  angle  included  by  the  secant  and  the  tangent  to  the  arc  at 
the  point  of  meeting. 

E.g.  in  the  figure  of  prop.  XXVIII,  OB  is  said  to  make  a  right  angle. 
with  the  circumference  EBA,  because  it  is  perpendicular  to  the  tangent 
at  B.  

Exercises.  318.  The  bisectors  of  the  interior  and  the  exterior  verti- 
cal angles  of  a  triangle  meet  the  circumscribed  circumference  in  the 
mid-points  of  the  arcs  into  which  the  base  divides  that  circumference, 
and  the  line  joining  those  points  is  the  diameter  which  bisects  the  base. 

319.  A  triangle  whose  angles  are,  respectively,  30°,  50°,  100°  is  inscribed 
in  a  circle  ;  the  bisectors  of  the  angles  meet  the  circumference  in  A,  B,  C. 
Find  the  number  of  degrees  in  the  angles  of  A  ABC. 

320.  The  three  sides  of  A  ABC  are,  respectively,  412  in.,  506  in., 
514  in.;  required  the  lengths  of  the  six  segments  formed  by  the  three 
points  of  tangency  of  the  inscribed  circle. 

321.  The  radii  of  two  concentric  circles  are  29  in.  and  36  in.,  respec-    -Ac- 
tively.    In  the  larger  circle  a  chord  is  drawn  tangent  to  the  smaller; 
required  its  length. 

322.  Two  circumferences  of  circles  of  radii  0.5  ft.  and  1.2  ft.  intersect 
so  that  the  tangents  drawn  at  their  point  of  intersection  are  perpendicu- 
lar to  each  other.     Required  the  distance  between  the  centers. 

323.  The  distance  between  the  centers  of  two  circles  of  radii  7  in.  and 
4  in.,  respectively,  is  8  in.  Required  the  length  of  their  common  tan- 
gent, between  the  points  of  tangency.     Is  there  more  than  one  answer  ? 

324.  The  distance  between  the  centers  of  two  circles  of  radii  327  in. 
and  115  in.,  respectively,  is  729  in.  Required  the  length  of  their  com- 
mon exterior  tangent,  between  the  points  of  tangency. 

325.  The  distance  between  the  centers  of  two  circles  is  165  in.  ;  the 
radii  are  62  in.  and  48  in.,  respectively.  Calculate,  correct  to  0.001,  the 
length  of  the  longest  line  parallel  to  the  center-line  and  30  in.  from  it, 
limited  by  the  circumferences. 

326.  Through  the  point  A,  6  in.  from  the  center  of  a  circle  of  radius 
4.5  in.,  two  tangents,  AT,  A  T ',  are  drawn.  Calculate  the  length  of  the 
chord  TT'  and  its  distance  from  the  center. 


X 


APPENDIX   TO    BOOK    III.  — METHODS. 


217.  The  student  has  already  been  informed  of  three  im- 
portant methods  of  attacking  a  proposition : 

(1)  By  Analysis  (§  113). 

(2)  By  Intersection  of  Loci  (I,  props.  XLIII,  XLIY). 

(3)  By  Reductio  ad  Absurdum  (§  74). 

He  is  now  prepared  to  discuss  these  somewhat  more  fully. 

218.  I.  Method  of  Analysis.  This  method,  first  found 
in  Euclid's  Geometry,  though  attributed  to  Plato,  may  be  thus 
described  :  Analysis  is  a  kind  of  inverted  solution;  it  assume* 
the  'proposition  proved,  considers  what  results  follow,  and  con- 
tinues to  trace  these  results  until  a  known  proposition  is  reached. 
It  then  seeks  to  reverse  the  process  and  to  give  the  usual,  or 
Synthetic,  proof. 

A  more  modern  form  of  analysis  is  sometimes  known  as  the 
Method  of  Successive  Substitutions.  In  this  the  student  sub- 
stitutes in  place  of  the  given  proposition  another  upon  which 
the  given  one  depends,  and  so  on  until  a  familiar  one  is  reached. 
The  student  reasons  somewhat  as  follows : 

1.  I  can  solve  A  if  I  can  solve  B. 

2.  And  I  can  solve  B  if  I  can  solve  C. 

3.  But  I  can  solve  C. 

Or  he  reasons  thus: 

1.  A  is  true  if  B  is  true. 

2.  And  B  is  true  if  C  is  true. 

3.  But  C  is  true. 

4.  Hence  A  and  B  are  true. 

162 


Sec.  218.] 


METHODS    OF  ATTACK. 


153 


Illustrative  Exercises.     1.   Through    a    given  point   to 
draw  a  line  to  make  equal  angles  with  two  Intersecting  lines. 


Analysis.  Suppose  x,  y  the  lines,  P  the  point,  and  I  the 
required  line ;  then,  in  the  figure,  Z.  c  =  Z.  a ■  +  Z  b ;  but  "•'  Z  a 
is  to  equal  Ah.  :.Z',  =  2Za;  .'.if  Zeis  bisected,  and 
a  line  is  drawn  through  P  parallel  to  this  bisector,  the  con- 
struction is  effected.  Now  that  the  method  is  discovered, 
give  the  solution  in  the  ordinary  way. 


2.  Tli rough  a  given  point  to  draw  a  line  such  that  the  seg- 
ments intercepted,  by  the  perpendiculars  let  fall  upon  it  from 
two  given  points  shall  be  equal. 


B 

\P 

x         D  \         D 

XA' 


Analysis.  Suppose  P  the  given  point  through  which  the 
line  x  is  to  be  drawn,  and  A  and  B  the  other  given  points ; 
then,  in  the  figure,  AD  and  BD'  _L  x,  and  DP  is  to  equal  PD'. 
Further,  if  AP  is  produced  to  meet  BD'  produced  at  A',  then 
A  DP  A  5£  A  D'PA'j  and  .'.  AP  =  PA'.  But  V  A  and  P  are 
given,  AP  can  be  drawn,  and  PA'  found  ;  .'.A'  can  be  found, 
and  .*.  A'B;  then  from  Pal  can  be  drawn  to  A'B,  and  the 
problem  is  solved.  Always  give  the  solution  in  the  ordinary 
way. 


154 


PLANE   GEOMETRY. 


[Bk.  III. 


3.  If  two  circles  are  tangent,  any  secant  drawn  through 
their  point  of  contact  cuts  off  segments  from  one  that  contain 
angles  equal  to  the  angles  in  the  segments  of  the  other. 


Analysis.     1.  Let  CD  be  the  common  tangent  to  ©  0,  0'  at 
their  point  of  contact  P.  Ill,  prop.  XXIII,  cor.  2 

2.  Then  an  Z  in  segment  A  =  an  Z  in  segment  A',  if 

Za  =  Z  a'.  Ill,  prop.  XIII 

3.  But  Za  =  Z  a'.  Prel.  prop.  VI 


. C V 

d^T      7\Y 


Exercises.  327.  To  construct  a  trap- 
ezoid, given  the  four  sides. 

Analysts.  Assume  the  figure  drawn. 
Then  if  d  is  moved  parallel  to  itself  and 
between  c  and  a,  to  the  position  FZ,  the 
A  XYZ  can   easily  be   constructed  (I, 

prop.  XXXIV).     The  process  may  now  be  reversed  and  the  trapezoid 
constructed. 

328.    To  place  a  line  so  that  its  extremities  shall  rest  upon  two  given 
circumferences,  the  line  being  equal  and 
parallel  to  another  line. 

Analysis.  If  0  and  O'  are  the  given 
circles,  &nd>AB  the  given  line,  and  if 
O  0'  is  moved  along  a  line  parallel  and 
equal  to  AB,  then  either  XY  or  X'Y' 
answers  the  conditions.  Hence  the 
process  may  be  reversed ;  first  describe 
O  0",  and  then  from  T,  T  draw  YX 
and  Y'X'  =  and  II  BA. 


Exs.  329-339.1  METHODS    OF  ATTACK,  155 


t 


EXERCISES. 

329.  Given  two  parallels,  AT.  X  Y'.  with  a  transversal  WZ  limited 
by  AT  and  X'Y' ;  also  two  points  A,  B.  not  between  the  parallels,  and 
on  opposite  sides  of  them.  Required  to  join  A  and  B  by  the  shortest 
broken  line  which  shall  have  MN, 

the  intercept  between  AFand  X'Y\ 
parallel  to  WZ. 

A  nalysis.  If  any  MN  in  the  figure 
is  moved  along  NB  parallel  to  its 
original  position,  until  N  coincides 
with  B  and  M  is  at  P,  then  AMXP  < 
AM2P    or    AMSP    (I,    prop.  VIII) ; 

hence  AM\N\B  is  the  shortest  broken  line.  Hence  the  process  may  be 
reversed  ;  first  draw  BP  II  and  =  ZW;  then  join  A  and  P,  thus  fixing  M\  : 
and  then  draw  M\N\  II  WZ. 

330.  Through  one  of  the  two  points  of  inter- 
section of  two  circumferences  to  draw  a  line  from 
which  the  two  circumferences  cut  off  chords  having 
a.  given  difference.  (The  projection  of  the  center- 
segment  on  the  required  line  equals  half  the  given 
difference  ;  hence  move  this  projection  to  the  position 

OA  ;  the  right-angled  A  OCA  can  now  be  constructed,  and  the  required 
line  will  be  parallel  to  OA.) 

331.  In  ex.  330,  show  that  if  the  two  chords  lie  on  opposite  sides  of  P. 
the  sum  replaces  the  difference. 

332.  In  a  given  circle  to  draw  a  chord  equal  and  parallel  to  a  given  line. 

333.  From  a  ship  two  known  points  are  seen  under  a  given  angle; 
the  ship  sails  a  given  distance  in  a  given  direction,  and  now  the  same  two 
points  are  seen  under  another  known  angle.  Find  the  positions  of  the 
ship.  (On  the  line  joining  the  known  points,  construct  segments  to 
contain  the  given  angles;    the  problem  then  reduces  to  ex.  328.) 

334.  Construct  a  trapezoid,  given  the  diagonals,  their  included  angle, 
and  the  sum  of  two  adjacent  sides. 

335.  To  construct  a  triangle  given  a  and  the  orthocenter.     /% 

336.  Also,  given  a  and  the  centroid.  *      xv  $A 

337.  To  draw  a  tangent  to  a  given  circle,  perpendicular  to  a  given  line. 

338.  To  construct  a  triangle,  ABC,  having  given  c.  Z  C,  and  the  foot 
of  the  perpendicular  from  C  to  c. 

339.  Find  the  locus  of  the  points  of  contact  of  tangents  drawn  from  a 
fixed  point  to  a  system  of  concentric  circles. 


* 


156  PLANE   GEOMETRY.  [Bk.  III. 

219.  II.  Method  of  Intersection  of  Loci.  This  method, 
adapted  chiefly  to  the  solution  of  problems,  has  already  been 
used  in  Book  I  (props.  XLIII,  XLIV).  So  long  as  it  is  known 
merely  that  a  point  is  on  one  line,  its  position  is  not  definitely 
known ;  but  if  it  is  known  that  the  point  is  also  on  another 
line,  its  position  may  be  uniquely  determined.  For  example, 
if  it  is  known  that  a  point  is  on  each  of  two  intersecting  lines, 
the  point  is  uniquely  determined  as  their  point  of  intersection ; 
but  if  the  point  is  on  a  straight  line  and  a  circumference  which 
the  line  intersects,  it  may  be  either  of  the  two  points  of  inter- 
section. 

For  convenience  of  reference  the  following  theorems  are 
stated,  and  will  be  referred  to  by  the  letters  prefixed : 

a.  The  locus  of  points  at  a  given  distance  from  a  given  point 
is  the  circumference  described  about  that  point  as  a  center,  with 
a  radius  equal  to  the  given  distance.     (§  127.) 

b.  The  locus  of  points  at  a  given  distance  from  a  given  line 
consists  of  a  pair  of  parallels  at  that  distance,  one  on  each  side 
of  the  fixed  line.     (§  129,  cor.  2.) 

c.  The  locus  of  points  equidistant  from  two  given  points  is 
the  perpjendicular  bisector  of  the  line  joining  them.      (§  128.) 

d.  The  locus  of  points  equidistant  from  two  given  lines  con- 
sists of  the  bisectors  of  their  included  angles  ;  if  the  lines  are 
parallel,  it  is  a  parallel  midway  between  them.      (§  129.) 

e.  The  locus  of  points  from  which  a  given  line  subtends  a 
given  angle  is  an  arc  subtended  by  that  chord.     (§  195,  cor.  3.) 

Abbreviations.  The  following  abbreviations  will  be  used  : 
In  the  triangle  ABC  the  altitudes  on  the  sides  a}  b,  c  will 
be  designated  by  hn,  hh,  hc,  respectively ;  the  corresponding 
medians  by  ma,  mb,  mc ;  the  corresponding  angle-bisectors 
terminated  by  a,  b,  c,  by  va,  vb,  vc;  the  radii  of  the  inscribed 
and  circumscribed  circles  by  r,  B,  respectively ;  the  radius  of 
the  escribed  circle  touching  a,  and  touching  b  and  c  produced, 
by  ra,  and  similarly  for  rb,  rc. 


Sec.  220.]  METHODS    OF  ATTACK.  157 

220.    Definition.     A  triangle  is  said  to  be   inscribed   in 

another  when  its  vertices  lie  respectively  on  the  sides  of  the 
other. 


Exercises.    340.    To  describe  a  circumference  with  a  given  radius,  and 
*     (1)  Passing  through  two  given  points.     (Combine  a  and  c.) 
X    (2)  Passing  through  one  given  point  and  touching  a  given  line,     (a,  b.) 

(3)  Passing  through  one  given  point  and  touching  a  given  circle,     (a. ) 

(4)  Touching  a  given  line  and  a  given  circle,     (a,  6.) 

(5)  Touching  two  given  circles,     (a.) 

341.    In  a  given  triangle  to  inscribe  a  triangle  with  two  of  its  sides 
given,  and  the  vertex  of  their  included  angle  given,     (a.) 

-V-       342.    To  describe  a  circumference  passing  through  a  given  point  and 
touching  a  given  line,  or  a  given  circle,  in  a  given  point,     (c.) 

-X      343.    On  a  given  circumference  to  find  a  point  having  a  given  distance 
from  a  given  line.    '(6.) 

f344.    On  a  given  line,  not  necessarily  straight,  to  find  a  point  equi 
distant  from  two  given  points,     (c.) 

t345.    Describe  a  circumference  touching  two  parallel  lines  and  passing 
through  a  given  point,     (d,  a.) 

346.  Find  a  point  from  which  two  given  line-segments  are  seen  under 
cf    (or  subtend)  given  angles,     (e.)     (Pothenot's  problem.) 

347.  Construct  the  triangle  ABC.  given  o,  //,„  ma. 

348.  Also,  given  ZA,  a,  h(l.     — 
—         349.    Also,  given  ZA,  a,  ma. 

*     350.    Also,  given  a,  hb,  hc. 
I    X    351.    Also,  given  Z  A,  ha,  va.     (First  construct  the  right-angled  tri- 
angle with  side  ha  and  hypotenuse  va.) 

352.  Also,  given  ha,  ma,  B.     (First  construct  the  right-angled  triangle 
with  side  hn  and  hypotenuse  ma  ;  then  find  the  circumcenter  by  a,  c.) 

353.  Also,  given  a,  R.  hb.     (First  construct  the  right-angled  triangle 
with  side  hb  and  hypotenuse  a  ;  then  find  the  circumcenter  by  a.) 

354.  Also,  given  c,  r,  Z  A  =  90° ;   c,  r,.  Z  A  =  90° ;   6.  r.  ZA  =  90°  ; 
or  6,  rc,  Z  A  =  90°. 

355.  Describe  two  circles  of  given  radii  /v  r2,  to  touch  one  another, 
and  to  touch  a  given  line  on  the  same  side  of  it. 


4 


158  PLANE   GEOMETRY.  [Bk.  III. 

MISCELLANEOUS  EXERCISES. 

356.  If  two  circumferences  intersect,  any  two  parallel  lines  drawn 
through  the  points  of  intersection  and  terminated  by  the  respective 
circumferences  are  equal. 

357.  If  the  center-segment  of  two  circles  is  (1)  greater  than,  (2)  equal 
to,  the  sum  of  the  two  radii,  the  circumferences  (1)  do  not  meet,  (2)  are 
tangent. 

358.  The  greatest  of  all  lines  joining  two  points,  one  on  each  of  two 
given  circumferences,  is  greater  than  the  center-segment  by  the  sum  of 
the  radii. 

359.  If  two  circles,  whose  centers  are  O,  (7,  are  tangent  at  P,  and  a 
line  through  P  cuts  the  circumferences  at  A,  A',  prove  that  OA  II  (/A '. 
Two  cases ;  external  and  internal  tangency.  Show  that  the  proposition 
is  true  for  any  number  of  circles. 

360.  Through  a  vertex  of  a  triangle  to  draw  a  straight  line  equally 
distant  from  the  other  vertices. 

361.  Describe  a  circle  of  given  radius  to  touch  two  given  lines.  Show 
that  a  solution  is,  in  general,  impossible  if  the  lines  are  parallel,  but  that 
otherwise  there  are  four  solutions. 

362.  From  what  two  points  in  the  plane  are  two  circles  seen  under 
equal  angles  ? 

363.  Given  an  equilateral  triangle,  ABC,  find  a  point  P  such  that  the 
circles  circumscribing  4  PBC,  PC  A,  PAB  are  all  equal. 

364.  To  divide  a  circle  into  two  segments  so  that  the  angle  contained 
in  one  shall  be  double  that  contained  in  the  other. 

365.  From  two  given  points  to  draw  lines  meeting  a  given  line  in  a 
point  and  making  equal  angles  with  that  line,  the  points  being  on  (1)  the 
same  side  of  the  given  line,  (2)  opposite  sides  of  the  given  line. 

366.  To  draw,  through  a  given  point,  a  secant  from  which  two  equal 
circumferences  shall  cut  off  equal  chords.  Discuss  the  number  of  solu- 
tions for  various  positions  of  the  given  point. 

367.  Through  one  of  the  points  of  intersection  of  two  circumferences 
to  draw  a  chord  of  one  circle  which  shall  be  bisected  by  the  circumference 
of  the  other. 

368.  Two  opposite  vertices  of  a  given  square  move  on  two  lines  at 
right  angles  to  each  other.  Find  the  locus  of  the  intersection  of  the 
diagonals. 

369.  Find  the  locus  of  the  intersection  of  two  lines  passing  through  two 
fixed  points  on  a  circumference  and  intercepting  an  arc  of  constant  length. 


BOOK   IV.  —  RATIO   AND   PROPORTION. 


1.     FUNDAMENTAL   PROPERTIES. 

221.  Introductory  Xote.  The  inference  was  drawn  in 
Book  II  (§  155)  that  a  relation  exists  between  algebra  and 
geometry  with  the  following  correspondence : 

Geometry.  Algebra. 

A  line-segment.  A  number. 

The  rectangle  of  two  line-segments.       The  product  of  two  numbers. 

And  as  it  was  assumed  that  a  straight  line  may  be  represented 
by  a  number,  so  it  may  be  assumed  that  any  other  geometric 
magnitude,  such  as  an  arc,  an  angle,  a  surface,  etc.,  may  be 
represented  by  a  number.  With  these  assumptions,  the  fun- 
damental properties  of  Batio  and  Proportion  may  be  proved 
either  "by  algebra  or  by  geometry,  as  may  be  most  convenient, 
the  proof  being  valid  for  both  of  these  subjects.  The  purely 
geometric  treatment  is  too  difficult  for  the  beginner. 

222.  Definitions.  To  measure  a  magnitude  is  to  find  how 
many  times  it  contains  another  magnitude  of  the  same  kind, 
called  the  unit  of  measure. 

223.  A  ratio  is  the  quotient  of  the  numerical  measure  of 
one  magnitude  divided  by  the  numerical  measure  of  another 
magnitude  of  the  same  kind. 

Tor  example,  the  ratio  of  a  line  8  ft.  long  to  one  16  ft.  long  is  T8?,  or  \  ; 
that  of  one  16  ft.  long  to  one  8  ft.  long  is  2. 

The  ratio  of  a  to  b  is  expressed  by  the  symbols  -  •  a :  6,  a/5,  or  a  -4-  b. 

If  the  ratio  r  =  r,  then  a  =  r  •  b. 
b 

159 


160  PLANE    GEOMETRY.  [Bk.  IV. 

224.  The  practical  method  of  finding  the  ratio  of  two 
magnitudes  is  to  measure  them,  and  to  divide  the  numerical 
result  of  one  measurement  by  that  of  the  other.  But  if  two 
line-segments  have  a  common  measure,  their  ratio  and  their 
common  measure  may  be  found  by  the  following  process : 

Let   AB   and    CD    be   the 
two  lines.  A« ■ ■ — ►-.b 

Apply  CD  as  often  as  pos-  F  ,  ,  ,p 
sible  to  AB,  and  suppose  that 

AB  =  2  CD  +  EB,  EB  <  CD. 
Similarly,  apply  EB  to  CD,  and  suppose  that 

CD  =  2EB  +  FD,  FD  <  EB. 
Similarly,  apply  FD  to  EB,  and  suppose  that 

EB  =  FD+  GB,   GB  <  FD. 
Similarly,  apply  GB  to  FD,  and  suppose  that 

FD  =  3  GB  with  no  remainder. 
Then  FD  =  3  GB. 

EB  =     FD  +  GB=    4  GB. 

CD  =  2EB  +  FD=    8  GB  +  3  GB  =  11  GB. 

AB  =  2  CD  +  EB  =  22  GB  +  4  GB  =  26  ££\ 

.*.  Gi?  is  a  common  measure,  and  the  ratio  of  CZ>  to  AB  is 
i£  by  definition  of  ratio. 

225.  Definitions.  Two  magnitudes  that  have  a  common 
measure  are  said  to  be  commensurable;  if  they  have  no  com- 
mon measure  they  are  said  to  be  incommensurable. 

For  example,  two  surfaces  having  areas  10  sq.  in.  and  15  sq.  in.  are 
said  to  be  commensurable,  there  being  the  common  measures  5  sq.  in., 
1  sq.  in.,  2.5  sq.  in.,  etc.  But  if  the  length  of  one  line  is  represented 
by  V2,  and  the  length  of  another  by  1,  then  there  is  no  common  measure, 
and  the  lines  are  said  to  be  incommensurable. 

A  ratio  may  therefore  be  an  integer,  or  a  fraction,  or  an  irrational 
number  such  as  V2. 


Prop.  I.]  RATIO  AND   PROPORTION.  161 

For  practical  purposes  all  magnitudes  may  be  looked  upon 
as  commensurable,  since  a  unit  of  measure  can  be  so  taken 
that  the  remainder  may  be  made  as  small  as  we  wish. 

226.  In  the  ratio  a  :  b,  a  and  b  are  called  the  terms  of  the 
ratio, — the  former,  a,  being  called  the  antecedent,  and  the 
latter,  b,  the  consequent. 

227.  If  the  ratio  a  :  b  equals  the  ratio  c  :  d,  the  four  terms 
are  said  to  form  a  proportion. 

The  four  terms  are  also  said  to  be  in  proportion.  The  terms 
a  and  b  are  also  said  to  he  proportional  to  c  and  d. 

(I        c 
This  equality  of   ratios  is  indicated  by  the  symbol  =,  e.g.,  t=^' 

a  :  b  =  c  :  d,  or  a/b  =  c/d,  read  "a  is  to  b  as  c  is  to  cZ.'1  Instead  of  the 
parallel  bars  (  =  ),  the  double  colon  (: :)  is  also  used  in  this  connection 
as  a  sign  of  equality,  the  proportion  being  written  a  :  b  :  :  c  :  d.  The 
double  colon  is  not.  however,  as  extensively  used  as  formerly. 

228.  The  first  and  last  terms  of  a  proportion  are  called  the 
extremes,  and  the  other  terms  the  means. 

Thus  in  the  proportion  2:3  =  6:  9,  3  and  6  are  the  means  and  2  and  9 
are  the  extremes. 


Proposition  I. 
229.    Theorem.     If  a  :  b  =  c  :  d,  then  ad  =  be. 

(Jb  C 

Proof.    From  j-  =  ;v  it  folloAvs,  by  multiplying  equals  by  bd, 

that  ad  =  be.  Ax.  6 
Hence 

If  four  numbers  are  in  pro-  If  four  lines  are  in  propor- 

portion,    the  product   of   the  tion,    the    rectangle    of    the 

means  equals  the  product  of  means  equals  the  rectangle  of 

the  extremes.  the  extremes. 


1G2 


PLANE    GEOMETRY. 


[Bk.  IV, 


Proposition  II. 
230.    Theorem.     If  ad  =  be,  then  a  :  b  =  c  :  d. 


Proof.    Divide  the  given  equals  by  bd. 

Hence 

If  the  product  of  two  num- 
bers equals  the  product  of  two 
other  numbers,  either  two  may 
be  made  the  means  and  the 
other  two  the  extremes  of  a 
proportion. 


Ax. 


If  the  rectangle  of  two  lines 
equals  the  rectangle  of  two 
other  lines,  either  two  may 
be  made  the  means  and  the 
other  two  the  extremes  of  a 
proportion. 


Proposition  III. 

-* 

231.    Theorem.     If  a  :  b  =  c  :  d,  then  (1)  a  :  c  =  b  :  d, 
(2)  d  :  b  =  c  :  a,  and  (3)  b  :  a  =  d  :  c. 


Proof.    1. 


and 


and 
2.  And 
3. 


Hence 

If  four  numbers  are  in  pro- 
portion, the  following  inter- 
changes may  be  made  :  (1)  the 
means,  (2)  the  extremes,  (3) 
each  antecedent  and  its  cor- 
responding consequent. 


Prop.  I 
Prop.  II 

Prop.  II 

Stepl 

Prop.  II 


If  four  magnitudes  are  in 
proportion,  the  following  in- 
terchanges may  be  made:  (1) 
the  means,  (2)  the  extremes, 
(3)  each  antecedent  and  its 
corresponding  consequent. 


ad  - 

=  be, 

a 
c 

b 

%7 

which 

proves 

(i), 

d 

b  = 

6 

=  —j 
a 

which 

proves 

(2)- 

be  = 

=  ad 

b 

a 

d 

—  — j 

c 

which 

proves 

(3). 

Prop.  IV.]  RATIO  AND  PROPORTION.  163 

Definitions.  The  proportion  a  :  c  =  b  :  d  is  often  spoken  of 
as  the  proportion  a  :  b  =  e  :  d  taken  by  alternation. 

The  proportion  b  :  a  =  d  :  e  is  also  spoken  of  as  the  propor- 
tion a:b  =  c:d  taken  by  inversion. 

Hence  prop.  Ill  may  be  stated:  If  four  magnitudes  are 
in  proportion  they  are  in  proportion  by  alternation  and  also 
by  inversion. 

But  to  take  a  proportion  by  alternation,  the  magnitudes  must  be  similar. 
Thus  1 2  :$4  =  $8  :|  16,  therefore,  by  alternation,  82  :  88  =  84  :  $16. 
But  the  proportion  82:84  =  8  days  :  16  days  cannot  be  taken  by  alter- 
nation, for  $2:8  days  =  84  :  16  days  means  nothing,  $2:8  days  not 
being  a  ratio  (§  223). 


Proposition  IV. 

232.    Theorem.     If  a  :  b  =  c  :  d, 

then  (1)  ka  :  b  =  kc  :  d, 
and  (2)  a  :  kb  =  c  :  kd. 

01  G 

Proof.    From  -  =  ->  it  follows,  by  multiplying  by  k, 

that  —  =  -7 j  which  proves  (1).  Ax.  6 

b         d  v  7 

a         c 

And  also,  —  =  —  >  by  dividing  by  /.',  which  proves  (2). 

nO         KCL 

Ax.  7 

Hence  if  four  magnitudes  are  in  proportion,  and  both  ante- 
cedents or  both  consequents  are  multiplied  by  the  same  num- 
ber, the  magnitudes  are  still  in  proportion. 

Corollary.  If  four  magnitudes  are  in  proportion,  and  all 
are  multiplied  by  the  same  number,  the  results  are  in  proportion. 

Note.     The  number  k  may  be  integral,  fractional,  or  irrational. 


164  PLANE   GEOMETRY.  [Bk.  IV. 

Proposition  V<. 
233.   Theorem.    If  a  :  b  =  c  :  d, 

then      (1)  a±b:b  =  c±d:d, 

(2)  a  ±  b  :  a  =  c  ±  cl  :  c, 

and       (3)    a±b:a=f:b  =  c±d:cq=d. 

Proof.    1.   From  t  =  v   it  follows 

6      a 

Axs.  2,  3 
which  proves  (1). 


Prop.  Ill 
Axs.  2,  3 


that 

or 

6             at 

2.  It  is 

also  true 

that 

b_d 

a      c 

and 

a              c 

or 

U±b       c±d 

which  proves  (2). 

to  V 

3.  Or,  by  subtracting  first, 


a 

a 

c  qp  d 
c 

a 

±b 

e  ±  d 

Axs.  3,  2 


and  .'. =- ;   bv  dividing   in  the   last  two 

a  if  b      c  ifd      J 

equations.  Ax.  7 

The  proportion  a  +  b  :  b  =  c  +  d  :  d  is  often  spoken  of  as 
the  proportion  a  :  b  =  c  :  d  taken  by  composition,  and  a  —  b  :  b 
=  c  —  d  :  d  as  the  same  proportion  taken  by  division,  and 
a±b:azpb  =  c±dicqpd  as  the  same  proportion  taken  by 
composition  and  division. 


Props.  VI,  VII.]         RATIO  AND  PROPORTION.  165 

Proposition*  VI. 

234.    Theorem.     If  ^  =  J-2-  =  ^  =  ,  the  terms  all  beinq 

\      b2      b3  y 

magnitudes  of  the  same  kind,  then 


a-.       a9 
-±  or  —  or 


bj  +  b2  + bj      b 


Proof.    1.  ^i  =  |l,  Prop.  Ill 

aud       ...fd^s^-iiJs.  p     v 

«2  ^2 

2.  ...^±?=fS  Prop.  Ill 

#1    +    62  #2 


3.  Then,  as  in  steps  1,  2, 

ax  +  a2  +  a3      a3 


Given 


&1    +  &2  +   &3         l>% 

and  so  on,  however  many  ratios  there  may  be. 


Proposition  VII. 
235.    Theorem.     If  a  :  b  =  c  :  d,  then  a2  :  b2  =  c2  :  d2. 

_  a      c  az      c 

Proof.  ,._  =  _,..._  =  _  Ax.6 

Hence 

If  four  numbers  are  in  pro-  If  four  lines  are  in  propor- 

portion,  their  squares  are  also  tion,  the  squares  on  those  lines 

in  proportion.  are  also  in  proportion. 

Corollary.     If  a :  b  =  c  :  d,  and  m  :  n  =  x  :  y,  then  am  :  bn 
=  ex :  dy. 


166  PLANE    GEOMETRY.  [Bk.  IV. 

Proposition  VIII. 

236.  Theorem,     a  :  b  =  ka  :  kb. 

Proof.  kab  =  kab,  or  a  •  kb  =  b  •  ka. 

.'.  a  :  b  =  ka  :  kb.  Prop.  11 

Note.     The  number  k  may  be  integral,  fractional,  or  irrational. 

237.  Definitions.  If  a  :  b  =  c  :  x,  x  is  called  the  fourth  pro- 
portional to  a,  b,  c. 

Corollaries.  1.  By  three  of  the  four  terms  of  a  proportion 
the  other  is  determined. 

For  if  a  :'b  =  c  :  x,  or  x  :  b  =  c  :  a,  or  c  :x  =  d  :  6,  etc.,  it  follows  that 
ax  ■=  be,  whence  x  =  bc/a,  a  fixed  number. 

2.  If  a  :  b  =  a  :  x,  then  b  =  x. 

For  if,  in  the  proof  of  cor.  1,  c  =  a,  then  6  =  x. 

238.  If,  in  a  proportion,  the  two  means  are  equal,  as  in 
a  :  x  =  x  :  b,  this  common  mean  is  called  the  mean  proportional, 
or  geometric  mean,  between  the  two  extremes. 

Corollaries. 

The  mean  proportional  be-  The    geometric     mean    be- 

tween two  numbers  equals  the  tween  tivo  lines  equals  the  side 
square  root  of  their  product.         of  that  square   which   equals 

their  rectangle. 

Because  the  number  representing  the  square  units  of  area 
of  a  rectangle  is  the  product  of  the  two  numbers  representing 
the  linear  units  in  two  adjacent  sides,  the  expression  product 
of  two  lines  is  often  used  for  rectangle  of  two  lines. 


Exercises.     370.    Find  a  mean  proportional  between  2  and  32. 

371.  Find  a  fourth  proportional  to  3,  7,  15. 

372.  What  number  must  be  added  to  each  of  the  numbers  2,  1,  5,  3,  to 
have  the  results  in  proportion  ? 


Secs.  239,  240.]  THE    THEORY   OF  LIMITS.  167 


2.     THE   THEORY   OF   LIMITS. 

239.    Definitions.     A  quantity  is  called  a  variable  if,  in  the 
course   of   the    same  investigation,  it   may  take  indefinitely 

many  values ;  on  the  other  hand,  a  quantity  is  called  a  con- 
stant if.  in  the  course  of  the  same  investigation,  it  keeps  the 
same  value. 

M,  M2     M3 


B 


E.g.  if  a  line  AB  is  bisected  at  Mt,  and  MXB  at  M2,  and  M2B  at  3f3, 

and  so  on,  and  if  x  represents  the  line  from  A  to  any  of  the  points  Jfj , 

M2 ,  then  £  is  a  variable,  but  AB  is  a  constant. 

It  is  customary,  as  in  algebra,  to  represent  variables  by  the  last  letters 
of  the  alphabet,  and  constants  by  the  first  letters. 

240.  If  a  variable  x  approaches  nearer  and  nearer  a  con- 
stant a.  so  that  the  difference  between  x  and  a  can  become 
and  remain  smaller  than  any  quantity  that  may  be  assigned. 
then  a  is  called  the  limit  of  x. 

E.g.  in  the  above  figure.  AB  is  the  limit  of  x. 

But  if  the  point  M  simply  slides  along  the  line,  passing  through  B, 
then,  although  the  difference  between  AM  and  AB.  or  x  and  a,  can 
become  smaller  than  any  quantity  which  may  be  assigned,  it  does  not 
remain  smaller,  for  when  M  passes  through  B  this  difference  increases. 
Hence  AB  is  not  then  the  limit  of  x. 

That  "  x  approaches  as  its  limit  a  n  is  indicated  by  the 
symbol  x  =  a. 

Corollary.  If  x  =  a,  then  a  —  x  is  a  variable  whose  limit 
is  zero  :  thai  is,  a  —  x  =  0. 

Although  the  variable  has  been  taken,  in  this  discussion,  as  increasing 
towards  its  limit,  it  may  also  be  taken  as  decreasing.  Thus  if  we  bisect 
a  line,  bisect  its  half,  and  continue  to  bisect  indefinitely,  the  variable 
segment  is  evidently  approaching  a  limit  zero. 


168  PLANE    GEOMETRY.  [Bk.  IV. 

Proposition  IX. 

241.  Theorem.  If,  while  approaching  their  respective  limits, 
tivo  variables  have  a  constant  ratio,  their  limits  have  that  same 
ratio. 


B  C 

_j j. U- 


D 


Given  X  and  X',  two  variables,  such  that  as  they  increase 
they  approach  their  respective  limits  AB,  or  L,  and 
AC,  or  L',  and  have  a  constant  ratio  r. 

To  prove    that  L  :  V  =  r,  or  that  X :  X'  =  L  :  X'. 

Proof.  If  the  ratio  X:  X'  is  not  equal  to  the  ratio  L  :  L', 
then  (1)  it  must  equal  the  ratio  of  L  to  something 
less  than  L',  or  (2)  it  must  equal  the  ratio  of  L  to 
something  greater  than  V. 

It  will  be  shown  that  both  of  these  suppositions  are  absurd. 

I.    To  show  that  (1)  is  absurd. 

1.  Suppose  X:X'  =  L:L'-DC. 
Then           '.'  X:X'=  r, 

.'.X=rX', 
and  L  =  r(V  -  DC).     §  223,  def.  ratio 

2.  Then  L-  X=  r(L'  -DC-  X').  Ax.  3 

3.  But  v  X'=L) 

.'.  L'  —  X'  can  become  as  small  as  we  please. 

"  "         "       less  than  DC, 

and  .'.  r(L'  —  X'  —  DC)  can  become  negative. 

4.  But  V  X>  Lj  .'.  L  —  X  cannot  become  negative. 

5.  .*.  step  2  is  absurd,  for  a  negative  quantity  cannot 

equal  one  not  negative. 


Prop.  IX.]  THE    THEORY    OF  LIMITS.  169 

II.    To  show  that  (2)  is  absurd. 

1.  Suppose  X:  X'  =  L  :  V  +  CD'. 

Then  £  -  X  =  /•  (X'  +  CD'  -  X'), 

as  in  step  2,  p.  168. 

2.  But  r(D'  4-  CD'  —  X')  cannot  become  less  than 
r  •  CD'. 

3.  And  L  —  X—  0,  because  L  is  the  limit  of  X. 

4.  .'.if  step  1  were  true,  a  quantity,  L  —  X,  which  can 
become  as  small  as  we  please,  would  equal  a  quan- 
tity not  less  than  r  •  CD',  which  is  absurd. 

The  proof  would  be  substantially  the  same  if  the  two 
variables  were  supposed  to  decrease  toward  a  limit. 

Corollaries.     1.    If,    while    approaching    their   respective 
limits,  two  variables  are  always  equal,  their  limits  are  equal. 
For  their  ratio  is  always  1. 

2.  If,  while  approaching  their  respective  limits,  two  variables 
have  a  constant  ratio,  and  one  of  them  is  always  greater  than 
the  other,  the  limit  of  the  first  is  greater  than  the  limit  of  the 
second. 

For  the  limits  have  the  same  ratio  as  the  variables. 


Exercises.     373.    If  a :  b  —  c :  d, 
prove  that  a2bd  +  b~c  +  be  =  ab'2c  +  abd  +  ad. 

1.  Since  be  =  ad,  Prop.  I 
the  equation  is  true  if        a-bd  +  b'2c  =  ab2c  +  abd. 

2.  Now  if  in  place  of  each  ad  we  put  be,  we  see  that  the  equation  is 
true  if  ab'2c  -f  b2c  —  ab2c  +  b2c. 

But  this  is  an  identity.     Hence  the  proof  is  complete. 

374.    lia:b  =  c:d,  


prove  that  a  —  c  :  b  —  d  =  V  a-  +  c-  :  V&2  +  d2. 


Also  that  Va2  +  c2  :  \lb2  +  d2  =  A/ac  +  -  :  \Jbd  +  —  ■ 

Also  that  a  +  mb  •  a  —  nb  =  c.  +  md  :  c  —  nd. 


170  PLANE   GEOMETRY.  [Bk.  IV 


3.     A   PENCIL   OF  LINES    CUT   BY   PARALLELS. 

242.  Definitions.  Through  a  point  any  number  of  lines  can 
be  passed.     Such  lines  are  said  to  form  a  pencil  of  lines. 

The  point  through  which  a  pencil  of  lines  passes  is  called 
the  vertex  of  the  pencil. 


A  

A  pencil  of  three  lines.  A  pencil  of  four  parallels. 

The  annexed  pencil  of  three  lines  is  named  "V  —  ABC." 

To  conform  to  the  idea  of  a  general  figure,  set  forth  in  §§  94,  95,  the 

word  pencil  is  also  applied  to  parallel  lines,  the  vertex  being  spoken  of  as 

"  at  infinity." 

243.  Definition.  Two  lines  are  said  to  be  divided  proportion- 
ally when  the  segments  of  the  one  have  the  same  ratio  as  the 
corresponding  segments  of  the  other. 


Exercises.     375.    If  a :  b  =  c :  d,  prove  that 

(1)  a  +  b  +  c  +  d:b  +  d  =  c  +  d:d. 

(2)  in  (a  +  mb)  :n(a  —  nb)  =  m(c  +  md) :  n(c  —  nd). 

(3)  a  (a  +  b  +  c  +  d)  =  (a  +  6)  (a  +  c). 

(4)  a?c  +  ad2 :  b2d  +  bd2  =  (a  +  c)3 :  (b  +  d)*. 
376.    If  b  is  a  mean  proportional  between  a  and  c,  prove  that 

a2  -  b'2  +  c2 


1        1        1 
P  ~  P  +  <* 


b*. 


377.  Show  that  there  is  no  finite  number  which,  when  added  to  each 
of  four  unequal  numbers  in  proportion,  will  make  the  resulting  sums  in 
proportion. 

378.  If  a  :b  =  c  :d,  and  u  :v  =  x  :y, 

prove  that  au  +  bv  :  au  —  bv  =  ex  -f  dy  -.  ex  —  dy. 


Prop.  X. 


PENCILS    OF  LINES. 


171 


Proposition  X. 

244.  Theorem.  The  segments  of  a  transversal  of  a  pencil 
of  parallels  are  'proportional  to  the  corresponding  segments 
of  any  other  transversal  of  the  same  pencil. 


I    1 

,/H 

1== 

n  m 

nil/. 

t— 

!/ 

, 

/ 

(I 

\ 

bJ/-~. 

n  m 

Given  the  pencil  of  parallels  P,  cutting  from  two  trans- 
versals T  and  T  the  segments  A,  B  and  C,  £>, 
respectively. 

To  prove    that  A  :  B  =  C  :  D. 

Proof.  1.  Suppose  A  and  B  divided  into  equal  segments  I, 
and  that  A  =  reZ,  while  B  =  ?i7. 

(In  the  figure,  n  =  6,  n'  =  4.) 
Then  if   ll's  to  P  are  drawn  from  the  points  of  divi- 
sion, C  is  the  sum  of  n  equal  segments  m,  and  D  is 
the  sum  of  n'  equal  segments  w. 

I,  prop.  XXVII,  cor.  1 

A  _  wZ  _  ?i       ?iw.  _  6' 
'""  ~B  =  nTl  =  n'  =  n^~i=~D' 


Whv  ? 


Note.    The  preceding  proof  assumes  that  -4.  and  B  are  commensurable. 
The  following  proof  is  valid  if  A  and  B  are  incommensurable. 


172  PLANE    GEOMETRY 

245.    Proof  for  incommensurable  case. 


[Bk.  IV 


1.  Suppose  A  divided  into  equal  segments  I,  and  that 

A  —  nl, 
while  B  =  n'l  +  some  remainder,  x,  such  that  x  <  /. 
Then  if  ll's  to  P  are  drawn  from  the  points  of  divi- 
sion, C  is  the  sum  of  n  equal  segments  m,  and  D  is 
the  sum  of  n'  equal  segments  m,  +  a  remainder  y 
such  that  y  <  m. 

2.  Then  Z^  lies  between  n'l  and  (w'  +  1)  /.  Step  1 


3. 


.'.  —  lies  between  — -  and -r— 

A  nl  nl 


4jt 
in  the  figure,  between  —  and 


(>/ 


) 


D  n'm  (n[  +  1)  m 

while    —  lies  between and L 

(  nm,  nm 

4.  .'.  —  and  —  both  lie  between  —  and 

A  C  n  n 


and       .'.  they  differ  by  less  than 
(In  the  figure,  by  less  than  \.) 


1 


Prop.  X.]  PENCILS    OF  LINES.  173 

5.   And  v  -   can  be  made  smaller  than  any  assumed 
n 

difference,  by  increasing  n}  .'.to  assume  any  differ- 
ence leads  to  an  absurdity. 

a  .    B      D        A      C  JJT 

b.  ■'A=r0T^==D'  PropI11 

Corollaries.     1,  A  line  parallel  to  one  side  of  a  triangle 

divides  the  other  two  sides  proportionally. 


For  in  the  figure,  if  BCO  is  the  triangle,  the  lines  OB.  OC  are  cut  by 
parallels.     Hence  BBX  : BxO  =  CCX  :  CxO. 

2.  The  corresponding  segments  of  the  lines  of  a  pencil  cut 
off  (from  the  vertex)  by  parallel  transversals  are  proportional. 

In  the    above   figure,    0.4  :  0AX  =  OB  :  0BX  =  OC  :  0C\  = by 

prop.  X. 

3.  The  segments  of  the  lines  of  a  pencil  cut  off  {from  the 
vertex)  by  parallel  transversals  are  proportional  to  the  corre- 
sponding segments  of  the  transversals. 

To  prove  that,  in  the  above  figure,  AB  :  AXBX  =  OA  :OAi=  OB  :  0BV 
Draw  through  Ax  a  line  II  to  OB  cutting  AB  at  X. 
Then  AB  :  XB  =  OAiOAx=  OB  :  0BX.  Prop.  X 

But  XB  -  AiBv  I,  prop.  XXIV 

4.  Parallel  transversals  are  divided  proportionally  by  the 
lines  of  a  pencil. 

To  prove  that,  in  the  above  figure,  AB  \BC  =  A\B\\  BXCX. 
By  cor.  3,  AB  :  AXBX  =  BO  :  Bx0  =  BC  :  BXCX.     Hence,  etc. 


174 


PLANE   GEOMETRY. 


[Bk.  IV. 


Proposition  XI. 

246.  Theorem.  A  line  can  be  divided,  internally  or  ex- 
ternally, into  segments  having  a  given  ratio,  except  that  if  it 
is  divided  externally  the  ratio  cannot  be  unity. 


Given 


Fig.  l.  Fig.  2. 

the  line  AB,  and  two  lines  s1}  s2  having  a  given  ratio. 


To  prove  that  AB  can  be  divided  in  the  ratio  Sj :  s2,  except  that 
in  the  case  of  external  division  sx  cannot  equal  s2. 

Proof.  1.  Suppose  AM  drawn  making,  with  AB,  an  angle 
<  180°;  that  AC  be  taken  =  s1}  and  CD  =  s2 ;  that 
DB  be  drawn,  and  CP  II  DB. 

2.  Then  AP  :  PB  =  sx :  s2,  as  required.       Prop.  X,  cor.  1 

3.  In  Pig.  2,  if  st  =  s2,  where  does  D  fall  ?  What  is 
then  the  relation  of  CP  to  AB  ?  Hence  show  that 
the  division  is  impossible  in  this  case. 

Corollary.  The  point  of  internal  division  is  unique ;  like- 
wise the  point  of  external  division. 

From  step  2,  AB  :  PB  =  Si  +  s2  :  s2,  AB,  Si  -f  s2,  and  s2,  all  being 
constants  ;  but  by  three  terms  of  a  proportion  the  fourth  is  determined. 
(§  237,  def.  of  4th  proportional,  cor.  1.) 

247.  Note.  Instead  of  saying  that  the  external  division,  if  the  ratio 
is  unity,  is  impossible,  it  is  often  said  that  the  point  of  division,  P,  is  at 
infinity. 

In  the  case  of  internal  division,  the  ratios  AP  :  PB  and  AC  :  CD  are 
evidently  positive ;  but  in  the  case  of  external  division  each  ratio  is 
evidently  negative  because  PB  and  CD  are  negative.  In  both  cases 
step  2  is  evidently  true. 


Prop.  XII.  1 


PENCILS    OF  LINES. 


175 


Proposition  XII. 

248.    Theorem.    A  line  which  divides  two  sides  of  a  triangle 
proportionally  is  parallel  to  the  third. 


Given 


the  triangle  ABC,  and  DE  so  drawn  that  AD  :  DC 
=  BE :  EC. 


To  prove    that  DE  II  AB. 

Proof.    1.   Suppose  DE  not  ii  AB,  but  that  DX  II  AB. 

Then  BX :  XC  =  AD  :  D  C.  Prop.  X,  cor.  1 

2.  But  this  is  impossible,  for  the  division  of  BC  in  the 
ratio  AD  :  DC  is  unique.  Prop.  XI,  cor. 

3.  .'.  DX  must  be  identical  with  DE,  and  DE  II  AB. 
The  proof  is  the  same  for  all  of  the  figures. 


Exercises.    379.    In  the  above  figures,  if  AD  :  DC  =  BE  :  EC  =  m  :  n,  ^  { 
and  if  the  line  through  A  and  E  cuts  the  line  through  B  and  D  at  P, 
then  prove  that  J.P  :  PE  =  BP  :  PD  =  m  +  n  :  n. 

380.  If  ex.  379  has  been  proved,  show  from  it  that  the  centroid  of  a 
triangle  divides  the  medians  in  the  ratio  of  2:1. 

381.  Prove  prop.  XI  on  the  following  figures  : 


n 


176  PLANE    GEOMETRY.  [Bk.  IV. 

Proposition  XIIL 

249.  Theorem.  If  any  angle  of  a  triangle  is  bisected,  in- 
ternally or  externally,  by  a  line  cutting  the  opposite  side, 
then  the  opposite  side  is  divided,  internally  or  externally, 
respectively,  in  the  ratio  of  the  other  sides  of  the  triangle. 


Given         A  ABC,  the  bisector  of  Z  C  meeting  AB  at  P. 
To  prove    that  AP  :  PB  =  AC  :  BC. 

Proof.    1.  Let  BE  II  PC,  meeting  AC  produced  at  E,  in  Fig.  1. 
Then    Z  EBC  =  Z  PCB  =  Z  ACP  =  Z  CEB. 

Given  ;  I.  prop.  XVII,  cor.  2 

2.  .'.BC=CE.  Why? 

3.  But  in  A  A  BE,  AP  :  PB  =  A  C :  CE,  Prop.  X,  cor.  1 
and  .'.  AP  :  PB  =  AC :  BC  Why  ? 
The  proof  for  Pig.  2  is  the  same  if  step  1  is  changed 
to  Z  CBE  =  ZBCP  =  Z  PCX  =  Z  BEC 

250.  Definition.  When  a  line  is  divided  internally  and 
externally  into  segments  having  the  same  ratio,  it  is  said  to 
be  divided  harmonically. 

If  the  internal  and  external  points  of  division  of  AB,  in  prop.  XIII, 
are  P  and  P'  then  AB  is  divided  harmonically  by  P  and  P'. 


Exercise.  382.  The  hypotenuse  of  a  ri^ht-ancled  triangle  is  divided 
harmonically  by  any  pair  of  lines  through  the  vertex  of  the  right  angle, 
making  equal  angles  with  one  of  its  arms. 


Sec.  251.] 


PENCILS    OF  LINES. 


177 


4,     A  PENCIL   CUT   BY   ANTIPARALLELS    OR   BY  A 
CIRCUMFERENCE. 

251.    Definitions.     If  a  pencil  of  two  lines   0  —  XY  is  cut 
by  two  parallel  lines  AB,  MX,  and  if  MX  revolves,  through  a 


straight  angle,  about  the  bisector  of  Z  XO  Y  as  an  axis,  falling 
in  the  position  AVBU  then  AB  and  AXBX  are  said  to  be  anti- 
parallel  to  each  other. 

OA  and  0AX  are  called  corresponding  segments  of  the  pencil, 
as  are  also  OB  and  0BX.  A  and  Ax  are  called  corresponding 
l>oints,  as  are  also  B  and  Bx. 

Corollary.  If  Z  A  —  Z  Ax,  in  the  above  figure,  then  AB 
and  AXBX  are  antiparallel  to  each  other. 


Exercises.  383.  From  P,  a  given  point  in  the  side  AB  of  A  ABC, 
draw  a  line  to  AC  produced  so  that  it  will  be  bisected  by  BC. 

384.  Investigate  ex.  383  when  P  is  on  AB  produced. 

385.  If  the  vertices  of  A  XYZ  lie  on  the  sides  of  A  abc  so  that  x  II  a, 
y  II  b,  z  II  c,  then  X,  F,  Z  bisect  «,  b,  c. 

386.  In  prop.  XIII,  suppose  Z  B  =  Z  A  ;  also,  suppose  ZB<Z^. 

387.  In  any  triangle  the  line  joining  the  feet  of  the  perpendiculars 
from  any  two  vertices  to  the  opposite  sides  is  antiparallel  to  the  thiru  side. 

388.  In  A  ABC,  suppose  that  a  ±  c,  and  the  bisectors  of  the  interior 
and  exterior  angles  at  C  meet  AB  at  Pi,  P2.  Prove  that  if  a  circum- 
ference passes  through  Pi,  P2,  and  C,  (1)  P1P2  is  the  diameter,  (2)  AC  is 
a  tangent. 


178 


PLANE    GEOMETRY. 


[Bic.  IV. 


Proposition  XIV. 

252.    Theorem.     If  a  pencil  of  two  lines  is  cut  by  two  anti- 
parallel  lines,  the  corresponding  segments  form  a  proportion. 


Given         the  pencil  0  —  XY,   cut  by  the  antiparallels  AB, 
AXBX,  A  and  Ax  being  corresponding  points. 

To  prove    that  OA  :  OAx  =  OB  :  OBx. 

Proof.    1.  Suppose  MN  the  parallel  to  AB  which,  revolving, 
fixed  AXBX. 

Then  OAx  =  OM,  and  OBx  =  OX.    Def .  antipar.  §  251 

2.   But  OA  :  OM=  OB  :  OX.  Prop.  X,  cor.  2 

and    .'.  OA:  OAx  =  OB  :  OBx.  Substitution 

Corollary.  If  two  antiparallels  cut  a  pencil  of  two  Hues. 
the  product  of  the  segments  of  one  line  equals  the  product  of 
the  segments  of  the  other. 

Why  ?     What  is  meant  by  "  product  of  two  segments  "  ? 


Exercises.     389.    In   the    above    figures.   AB  :  ^4i7>!  =  OA  :  OA 
OB-.OBx.    (Prop.   X,  cor.  3,  etc.) 

390.  In  the  above  figures,  if  ^4i  coincides  with  B.  and  if  OB  =  b, 
OA  =  a,  OBx  =  bx,  then  62  =  abx. 

391.  If  from  the  vertex  of  a  right-angled  triangle  a  perpendicular  p  is 
drawn  cutting  the  hypotenuse  c  into  two  segments  x,  y,  adjacent  to  sides 
a,  6,  respectively,  then  (1)  a  and  p  are  antiparallels  of  the  pencil  6,  c ; 
(2)  a  is  a  mean  proportional  between  c  and  x ;  (3)  p  is  a  mean  propor- 
tional between  x  and  y ;  (4)  62  =  cy,  a2  =  ex,  and  .-.  a2  +  b2  =  c  (x  +  y) 
=  c2.      (Thus  a  new  proof  is  found  for  the  Pythagorean  proposition.) 


,=  * 


Prop.  XV.] 


PENCILS    OF  LINES. 


179 


Proposition  XV. 

253.  Theorem.  If  a  pencil  of  lines  cuts  a  circumference, 
the  product  of  the  two  segments  from  the  vertex  is  constant, 
whichever  line  is  taken. 

A, 
A, 


Fig.  1.— The  point  O  on 
the  chord. 


Fig.  2.  — The  point  O  on 
the  chord  produced. 


Given         ABX  and  AXB,  two  chords,  each  divided  at  0  into 
two  segments. 

To  prove    that  AO  •  OBx  =  AxO  •  OB. 

Proof.    1.  Suppose  AB,  AXBX  drawn. 

Then  £A  =  £  Av  Why  ? 

2.  .'.  AB  and  AXBX  are  antiparallel,  §  251,  cor. 

and        .'.  AO  ■  OBx  =  AxO  ■  OB.        Prop.  XIV,  cor. 

The  proposition  is  entirely  general  and .  should  be  proved  for  the  fol- 
lowing cases. 


Fig.  3.  —  The  point  O  at 
the  end  of  the  chord. 


FIG.  4.  —  Chord  AXB 
becomes  zero. 


Fig.  5.  —Chord  ABY  also 
becomes  zero. 


Corollary.  The  tangent  from  the  vertex  of  a  pencil  to  a 
circumference  is  a  mean  proportional  between  the  two  segments 
of  any  other  line  of  the  pencil . 

In  Fig.  4,  AO  ■  BxO=AxO  •  BO=B02.    Therefore  AO  :  BO=BO  :  BxO. 


W' 


180  PLANE   GEOMETRY.  [Bk.  IV. 

Proposition  XVI. 

254.  Theorem.  In  the  same  circle  or  in  equal  circles 
central  angles  are  proportional  to  the  arcs  on  which  they 
stand. 


Given  A  and  B,  two  central  angles  standing  on  arcs  C  and 

D,  respectively. 

To  prove    that  A:  B  =  C :  D. 

Proof.    1.  If  A  and  B  are  in  different   circles,  they  may  be 
placed  in  the  relative  positions  shown  in  the  figure. 

§  108,  def.  O,  cor.  2 
Suppose  A  and  B  divided  into  equal  A  x, 
and  suppose     A  =  nx,  and  B  =  n'x. 
(In  the  figure,  n  =  6,  nf  =  4.) 
2.  Then  C  is  divided  into  n  equal  arcs  y, 

"     y.     Ill,  prop.  I 


and 

D 

«    »' 

3. 

^4  _  7103 

w        ny  _  6' 

.  a     c 

"  B      I) 

Why  ? 

Why  ? 

Note.  The  above  proof  assumes  that  A  and  B  are  commensurable, 
and  hence  that  they  can  be  divided  into  equal  angles  x.  The  proof  on 
p.  181  is  valid  if  A  and  B  are  incommensurable. 


Skcs.  255,  256.]  PENCILS    OF  LINES. 

255.    Proof  for  incommensurable  case. 


181 


1.  Suppose  A  divided  into  equal  A  x,  and  suppose 
A  =  nx,  while  B  =  n'x  +  some  remainder  w,  such 
that  w  <  x. 

Then  C  is  divided  into  u  equal  arcs  y,  and  D  is  the 
sum  of  n'  equal  arcs  y  +  a  remainder  .?,  such  that 
z  <  y. 


2.  Then  B  lies  between  n'x  and  (»*'  +  1)  x, 
and  D  lies  between  n'y  and  (V  +  1)  y. 

B       ,  2>  v    ^   '.    ,  «'         '»'  + 1 

3.  .*..—  and  —  both  he  between  —  and 

AC  n  n 

(In  the  figure,  between  |  and  f.) 

J)  T)  1 

And  .*.  —  and  —  differ  by  less  than  -■ 
AC-  n 


Why  ? 
Why  ? 

Why  ? 


Why  ? 


4,  And  '.'  -  can  be  made  smaller  than  any  assumed 

n 

difference,  by  increasing  n. 

.'.  to  assume  any  difference  leads  to  an  absurdity. 

B      D      .  AC 

5.  .-._  =  -,  whence -  =  -• 

Corollary.     In  the  same  circle  or  in  equal  circles  sectors 
are  proportion"!  to  their  angles  or  to  their  arcs. 

256.    This  proposition  is  often  stated, 

A  central  angle  is  measured  by  its  intercepted  arc.    See  §  180. 


,v-v 


*3 


182 


PLANE    GEOMETRY. 


[Bk.  IV. 


5.     SIMILAR   FIGURES. 


257.  Definitions.  We  have  (§  59)  roughly  defined  similar 
figures  as  figures  having  the  same  shape.  But  this  is  unsatis- 
factory because  the  word  shape  is  not  defined.  We  therefore 
proceed  scientifically  to  define 

1.  Similar  systems  of  jioints,  and  then 

2.  Similar  figures. 

Two  systems  of  points,  Ax,  Bx,  C1}  and  A2,  B2,  C2,  ,  are 

said  to  be  similar  when  they  can  be  so  placed  that  all  lines, 

AXA2,  B^B2,    CxC2,  ,  joining   corresponding   points  form  a 

pencil  whose  vertex,  0,  divides  each  line  into  segments  having 
a  constant  ratio  r. 


In  the  figure,  OAx  :  OA2  =  OBx  :  OB2  = 


258.    Two  figures  are  said  to  be  similar  when  their  systems 
of  points  are  similar. 

The  symbol  of  similarity  w*,  already  mentioned,  is  due  to  Leibnitz. 
It  is  derived  from  the  letter  S. 

The  following  are  illustrations  of  similar  figures  involving 
circles: 


Concentric  circles. 


Any  circles. 


Secs.  259-261.] 


SIMILAR    FIGURES. 


183 


The  following  are  illustrations  of  similar  rectilinear  figures 
A2 


v 


Any  line-segments.  Four  similar  triangles.        Three  similar  quadrilaterals. 

259.  When  two  similar  figures  are  so  placed  that  lines 
through  their  corresponding  points  form  a  pencil,  they  are 
said  to  be  placed  in  perspective,  and  the  vertex  of  that  pencil 
is  called  their  center  of  similitude. 

The  figures  above  and  on  p.  182  are  placed  in  perspective,  and  in  each 
case  0  is  the  center  of  similitude. 

Two  similar  figures  may  evidently  be  so  placed  that  the  center  of 
similitude  will  fall  within  both,  or  between  them,  or  on  the  same  side  of 
both,  as  is  seen  in  the  above  illustrations. 

260.  Two  systems  of  points,  A1}  Blf  Cx,  and  A.2,  B2,  C.2, 

,  are  said  to  be  symmetric  with  respect  to  a  center  0  when 

all  lines,  A^42,  BXB2,  CXC2,  ,  are  bisected  by  0. 


261.  Two  figures  are  said  to  be  symmetric  with  respect  to  a 
center  when  their  systems  of  points  are  symmetric  -with  respect 
to  that  center. 

E.g.  in  the  figure,  A  AxBiCi,  A2B2C2  are  symmetric  with  respect  to  0. 


184 


PLANE    GEOMETRY. 


[Bk.  IV. 


262.  In  similar  figures,  if  the  ratio,  r,  known  as  the  ratio  of 
similitude,  is  1,  the  figures  are  evidently  symmetric  with  respect 
to  a  center.  Hence  Central  Symmetry  is  a  special  case  of 
Similar  Figures  in  Perspective. 

The  term  Center  of  Similitude  is  due  to  Euler. 


Corollaries.     1.    Congruent  figures  are  similar. 

For  if  made  to  coincide,  any  point  in  their  plane  is  evidently  a  center 
of  similitude,  the  ratio  of  similitude  being  1.  Or,  they  may  be  placed  in 
a  position  of  central  symmetry. 

2.  To  any  ])oint  in  a  system  there  is  one  and  only  one  corre- 
sponding point  of  a  similar  system  with  respect  to  a  given 
center.  * 

If  A\  and  A%  are  corresponding 
points  in  two  similar  systems  in  per- 
spective, and  0  is  the  center  of  simili- 
tude, then  every  point  Pi  on  OAi  has 
a  unique  corresponding  point  P2  on 
OA2. 

For  OAl:OA2  =  OP^  :  OP2, 

.-.  0P2  is  unique.  §  237,  cor.  1 


Exercises.  392.  What  is  the  limit  of  1/x  as  x  increases  indefinitely  ? 
of  1/(1  +  x)  as  x  =  0  ?   as  x  =  1  ? 

393.  In  A  ABC,  P  is  any  point  in  AB,  and  Q  is  such  a  point  in  CA 
that  CQ  —  PB  ;  if  PQ  and  BC,  produced  if  necessary,  meet  at  X,  prove 
that  CA  :  AB  =  PX  :  QX.     (From  P  draw  a  line  \\  AC.) 

394.  In  the  annexed  figure  of  a  "Diagonal  Scale,"  AB 
is  1  centimeter.  Show  how,  by  means  of  the  scale  and  a 
pair  of  dividers,  to  lay  off  1  millimeter,  0.5  millimeter,  0.3 
millimeter,  etc.  On  what  proposition  or  corollary  does  this 
measurement  of  fractions  of  a  millimeter  depend  ? 

/"  395.  A  BCD  is  a  parallelogram  ;  from  A  a  line  is  drawn  cutting  BD 
I  in  E,  BC  in  F,  and  DC  produced  in  G.  Prove  that  AE  is  a  mean  pro- 
j    portional  between  EF  and  EG. 

396.  ABC  is  a  triangle,  and  through  D,  any  point  in  c,  DE  is  drawn 
II  a  to  meet  b  in  E  ;  through  C,  CF  is  drawn  II  EB  to  meet  c  produced  in 
F.     Prove  that  AB  is  a  mean  proportional  between  AD  and  A F. 


V 

1 

6  ■ 

4  ' 
.1  ' 

:• 
l  - 


Prop.  XVII.]  SIMILAR   FIGURES.  185 

Proposition  XVII. 

263.    Theorem.     Tivo  triangles  are  similar  if  they  have  two 
angles  of  the  one  equal  to  two  angles  of  the  other,  respectively. 


Given         the  A  AXBXCX,  A2B2C2,  with  ZAl  =  Z.A2,  ZC1  = 
ZC2. 


To  prove    that  A  AXBX C1  ^  A  A2B2C. 


Proof.    1.  Place  one  A  on  the  other  so  that  Z  C\  coincides  with 
Z  C2,  as  at  0,  OA2  falling  on  OAx.     Let  OX2Xx  be 
any  line  through  0,  cutting  A2B2  at  X2,  and  AXBX 
at  Xx. 
.       Then  \-  Z.A2  =  ZAX, 

.'.  A2B2  II  AXBX.         I,  prop.  XVI,  cor.  1 

2.   .'.  OAx  :  OA2  =  OBx  :  OB2  =  OXx  :  OX2  = =  r. 

Prop.  X,  cor.  2 

V      3.  And  all  points  on   OAx  and   OBx  have  their  corre- 
sponding points  on  OA2  and  OB2,  respectively. 
V  §  262,  cor.  2 

4.  .'.  the  A  are  similar,  0  being  the  center  of  simili- 
tude. §  258 

Corollaries.    1.  Mutually  equiangular  triangles  are  similar. 

2.  If  two  triangles  have  the  sides  of  the  one  respectively 
parallel  or  perpendicular  to  the  sides  of  the  other,  they  are 
similar. 

For  by  §  86,  cor.  5,  they  can  be  proved  to  be  mutually  equiangular. 


y 


186  PLANE   GEOMETRY.  [Bk.  IV. 

Proposition  XVIII. 

264.  Theorem.  If  two  triangles  have  one  angle  of  the  one 
equal  to  one  angle  of  the  other,  and  the  including  sides  pro- 
portional, the  triangles  are  similar. 


Given         A  AXBXC^  A2B2C2,  such  that  Z  C\  =  Z  C2  and  ax :  a2 

=  h  :  b2. 

To  prove    that  A  AXBX  C\  v-~  A  A2B2  C2. 

Proof.    1.  v  Z  C2  =  Z  Cu  AA2B2C2  can  be  placed  on  AA1B1C1 
so  that  C2  falls  at  C1?  i?2  on  a1?  and  J2  on  Z>P 

2.  Then     V  C^2 :  CXAX  =  C\B2 :  C^l5 

.-.  A2B2  II  J^.  Props.  XII,  V 

3.  .'.A  AlB1Cl  and  A  A2B2C1  are  mutually  equiangular. 

I,  prop.  XVII,  cor.  2 

4.  .'.  AA^iCi  ^  AA2B2C\  and  its  congruent  AA2B2C2. 

Prop.  XVII,  cor.  1 

Exercises.  397.  ABC,  DBA  are  two  triangles  with  a  common  side 
AB.  If  P  is  any  point  on  AB,  and  PX  II  AC,  and  PF  II  AD,  meeting 
BC  and  PD  in  X  and  F,  respectively,  prove  that  A  YBX  -~  A  DPC. 

398.  ABCD  is  a  quadrilateral.  Prove  that  if  the  bisectors  of  A  A,  C 
meet  on  diagonal  BD,  then  the  bisectors  of  AB,  D  will  meet  on  diago- 
nal AC. 

399.  Construct  a  triangle,  having  given  the  base,  the  vertical  angle, 
and  the  ratio  of  the  remaining  sides.    (Intersection  of  loci  and  prop.  XIII.) 

400.  In  A  ABC,  CM  is  a  median  ;  A  BMC,  CM  A  are  bisected  by 
lines  meeting  a  and  b  in  B  and  Q,  respectively.     Prove  that  QR  II  AB. 


Prop.  XIX. 


SIMILAR   FIGURES. 


18' 


Proposition  XIX. 

265.    Theorem.     If  tivo  triangles  have   their  sides  propoi 
tional,  they  are  similar. 

C,  Cs 


Given  AA-^Bi^C^  A2B2C2  such  that  ax  :  a,  —  bx  :  I>2  =  cx  :  c2° 

To  prove    that  A  AlBlC1  ^  A  A2B2C2. 

Proof.    1.   On  CXAX,  C\BX  lay  off  C\X  =  b2,  and  CXY  =  a2,  and 
draw  XY. 


Then 
o 

3.  But 
and 

4. 
and 

5. 
and 


'.'  a1:a2  =  b1:  b2,  and  Z.  C\  =  Z.  C\, 

'.  A  XYC\  ^  A  AXBXCX.  Prop.  XVIII 

ax :  a2  =  Cj  :  r27  Given 

ax  :  a2  =  c,  :  A'3~.  Prop.  X,  cor.  3 

.'.  Cl  :r2  =  c,  :J7,  Why? 

A  AXBX d  —  A  A2B2C2. 


I.  prop.  XII 
Steps  2,  5 


Exercises.  401.  The  product  of  the  two  segments  of  any  chord 
drawn  through  a  given  point  within  a  circle  equals  the  square  of  half 
the  shortest  chord  that  can  be  drawn  through  that  point. 

402.  If  P  is  a  point  on  AB  produced,  the  tangents  fiom  P  to  all 
circumferences  through  A  and  B  are  equal,  and  hence  such  points  are 
concyclic. 

403.  If  from  any  point  P  on  the  side  CA  of  a  right-angled  triangle 
ABC,  PQ  is  drawn  perpendicular  to  the  hypotenuse  AB  at  Q,  then 
AP  •  A  C  =  AQ  ■  AB.  Suppose  P  to  be  taken  (1)  at  C  ;  (2)  at  A  ;  (3)  on 
A  C  produced. 


h' 


188 


PLANE   GEOMETRY. 


[Bk.  IV. 


Proposition  XX. 

266.    Theorem.     Similar  triangles  have  their  corresponding 
sides  proportional  and  their  corresponding  angles  equal. 


Given         two  similar  triangles,  AXBXCX,  A2B2C2,  Ax  correspond- 
ing to  A2,  Bx  to  B2,  Cx  to  C2. 

To  prove    that  AXBX :  A2B2  =  BXCX:  B2C2  = and  that  ZBX 

=  Z-B2 

Proof.    1.  Suppose  the  A  placed  in  perspective. 

Then  OAx :  OA2  =  OBx :  OB2=  OCx:  OC2.         §  258 

2.  .".  AXBX  II  A2B2,  and  so  for  other  sides. 

Props.  XII,  V 

3.  .-.  Z  OBxAx  =  ^  OB2A2,  and  Z  CxBxO  =  Z  C2B20. 

I,  prop.  XVII,  cor.  2 

4.  .'./]>!  =  Z7>2,  and  so  for  other  angles.      Ax.  2 

5.  Also,  OB,  :  OB2  =  AXBX :  A2B2, 

=  BXCX:  B2C2.  Prop.  X,  cor.  3 

6.  .*.  AXBX  :  A2B2  =  BXCX  :  B2C2,  and  so  for  other  sides. 
Note.     This  is  the  converse  of  props.  XVII,  XIX. 

Corollaries.     1.  The  corresponding  altitudes  of  two  similar 
triangles  have  the  same  ratio  as  any  two  corresj»> mling  sides. 
Why? 

2.    The  corresponding  sides  of  similar  triangles  are  opposite 
the  equal  angles. 

In  what  step  is  this  proved  ? 


Secs.  267, 268.]  SIMILAR   FIGURES.  189 

267.  Summary  of  Propositions  concerning  Similar  Triangles. 
Two  triangles  are  similar  if 

1.  (a)  Two  angles  of  the  one  equal  two  angles  of  the  other. 

Prop.  XVII 

(b)  They  are  mutually  equiangular.      Prop.  XVII,  cor.  1 

(c)  The  sides  of  the  one  are  parallel  to  the  sides  of  the 

other.  Prop.  XVII,  cor.  2 

(d)  The  sides  of  the  one  are  perpendicular  to  the  sides  of 

the  other.  Prop.  XVII,  cor.  2 

2.  One  angle  of  the  one  equals  one  angle  of  the  other  and  the 

including  sides  are  proportional.  Prop.  XVIII 

3.  Their  corresponding  sides  are  proportional.       Prop.  XIX 

If  two  triangles  are  similar, 

1.  They  are  mutually  equiangular.  Prop.  XX 

2.  Their  corresponding  sides  are  proportional.         Prop.  XX 

3.  Their  corresponding  altitudes   are  proportional  to  their 

corresponding  sides.  Prop.  XX,  cor.  1 

268.  It  should  further  be  observed  that,  in  general, 
Three  conditions  determine  congruence.  (See  §  90.) 
Two  conditions  determine  similarity. 

For  these  conditions  are 

1.  Two  angles  equal.     (Prop.  XVII.) 

2.  One  angle  and  one  ratio.     (Prop.  XVIII.) 

3.  Two  ratios  ;   for  if  the  sides  are  a,  b,  c,   and  a',  b',  c',   then  if 

-==—-,  and  -  =  — ,  the  A  are  similar,  since  -  must  also  equal  —  * 
b      b'  c      c'  c  c' 


Exercises.  404.  If  X  is  any  point  in  the  side  a,  or  a  produced,  of 
A  ABC,  and  if  r&  and  rc  are  the  radii  of  circles  circumscribed  about 
A  ABX  and  A  AXC,  respectively,  then  rh:rc  —  c\b.  (Join  the  centers 
and  prove  two  triangles  similar.) 

405.  If  one  of  the  parallel  sides  of  a  trapezoid  is  double  the  other, 
prove  that  the  diagonals  intersect  one  another  in  a  point  of  trisection. 


190  PLANE   GEOMETRY.  [Bk.  IV. 

Proposition  XXI. 

269.  Theorem.  If  two  polygons  are  mutually  equiangular 
and  have  their  corresponding  sides  proportional,  they  are 
similar. 

B,Z ^Z ^^> 

Given         two  polygons,  AXBXCX and  A2B2C2 ,  such  that 

ZAX  =  ZA2,  ZBX  =  ZB2,  , 

and       AXBX :  A2B2  =  BXCX :  B2C2  =  

To  prove    that         J^Ci ^  A,  B2  C2 

Proof.  1.  Place  A2B2  II  AXBX.  Then  v  the  A  of  one  polygon 
==  the  corresponding  A  of  the  other,  the  remaining 
sides  may  be  made  parallel  respectively. 

I,  prop.  XVII,  cor.  5 

2.  If    AXBX  >  A2B2,   then    BXCX  >  B2C2,  ,  because 

the  ratios  are  equal. 

3.  Draw  AXA2,  BXB2,  Then  AXBXB2A2  is  not  a  O; 

also  BXC\C2B2,  etc.;  and  AXA2  meets  BXB2  as  at  0, 
BXB2  meets  C^CVas  at  0',  etc.  I,  prop.  XXIV 

4.  But  Bx 0' :  B20'  =  BXCX:  B2C2 

=  AXBX :  A2B2  =  BxO:  B20, 

»  Prop.  X,  cor.  3 

which/is  impossible  unless  0  and  0'  coincide. 

Prop.  XI,  cor. 

5.  .'.  the  two  figures  are  similar,  and  (>  is  the  center 
of  similitude.  §  258 


Prop.  XXL]  SIMILAR   FIGURES.  191 

In  step  2,  if  AXBX  =  A2B2,  then  BXCX  =  B2C2, ,  and  the 

polygons  are  congruent  and  therefore  similar.         §  262.  cor.  1 

Corollaries.  1.  If  two  polygons  are  similar,  titer/  are 
mutually  equiangular  and  their  corresponding  sides  are  pro- 
portional. 

For  if  placed  in  perspective  as  on  p.  190, 

1.  OAx :  OA2  =  OBx  :  OB2.  §  258 

2.  .-.  AXBX II  A2B2,  and  so  for  other  sides.  Prop.  XII 

3.  .-.   ZL'i  =  Z  B2,  and  so  for  other  angles.         I,  prop.  XVII,  cor.  5 

4.  Also  AXBX  :  A2B2  =  BxO  :  B20  =  BXCX  :  B2C2  = Prop.  X,  cor.  3 

2.  Polygons  similar  to  the  same  polygon  are  similar  to  each 
other. 

For  they  have  angles  equal  to  those  of  the  third  polygon,  and  the 
ratios  of  their  sides  equal  the  ratios  of  the  sides  of  the  third  polygon. 

3.  The  perimeters  of  similar  polygons  have  the  same  ratio 
as  the  corresponding  sides. 

For  by  cor.  1,  AXBX :  A2B2  =  BXCX  :  B2C2  = =  r.    .-.  AXBX  +  BXCX 

+ :A2B2  +  B2C2  +  =  r.     (Why  ?) 

4.  Two  similar  polygons  can  be  divided  into  the  same  num- 
ber of  triangles  similar  each  to  each,  and  similarly  placed. 

For  0  and  (X  coincide,  and  the  figures  can  be  placed  having  0  within 
each.     The  triangles  AxOBx,  A2OB2  are  then  similar,  by  prop.  XVII. 


Exercises.  406.  If  from  a  point  outside  a  circle  a  pair  of  tangents 
and  a  secant  are  drawn,  the  quadrilateral  formed  by  joining  in  succes- 
sion the  four  points  thus  determined  on  the  circumference  has  the  rect- 
angles of  its  opposite  sides  equal. 

407.  AB  is  a  diameter,  and  from  A  a  line  is  drawn  to  cut  the  circum- 
ference in  C  and  the  tangent  from  B  in  D.  Prove  that  the  diameter  is 
the  mean  proportional  between  A  C  and  AD. 

408.  In  O  ABCD.  P,  Q  are  points  in  a  line  parallel  to  AB  ;  PA  and 
QB  meet  at  R,  and  PB  and  QC  meet  at  S.     Prove  that  RS  II  AB. 

409.  Chords  AB,  CB  are  produced  to  meet  at  P,  and  PF  is  drawn 
parallel  to  BA  to  meet  CB  produced  in  F.  Prove  that  PF  is  the  mean 
proportional  between  FB  and  FC. 


192  PLANE    GEOMETRY.  [Bk.  IV. 

Proposition  XXII. 

270.  Theorem.  In  a  right-angled  triangle  the  perpendicu- 
lar from  the  vertex  of  the  right  angle  to  the  hypotenuse 
divides  the  triangle  into  two  triangles  which  are  similar  to 
the  whole  and  to  each  other. 


D 

Given         A  ABC,  with  Z  C  a  right  angle,  and  CD  _L  AB. 
To  prove    that     (l)AiCD^A  ABC. 

(2)  A  CBD  ~  A  ABC. 

(3)  AACD^A  CBD. 

Proof.    1.  ' .  •  Z  CD  A  =  ZACB,  Why  ? 

and  Z  A  =  Z  A, 

.'.AACD^AABC,  which  proves  (1). 

Prop.  XVII 

2.  Similarly  A  CBD  ~  A  ABC,  which  proves  (2). 

Prop.  XVII 

3.  .*.  A  ACD  ~  A  CBD,  which  proves  (3). 

Prop.  XXI,  cor.  2 

Corollaries.  1.  Either  side  of  a  right-angled  triangle  is 
the  mean  proportional  between  the  hypotenuse  and  its  segment 
adjacent  to  that  side. 

For  from  step  1,  AB  :  AC  =  AC :  AD  ;  and  from  2,  AB.BC  =  BC-.DB. 

2.  The  perpendicular  from  the  vertex  of  the  right  angle  to 
the  hypotenuse  is  the  mean  proportional  between  the  segments 
of  the  hypotenuse. 

For  from  step  3,  AD  :  CD  =  CD  :  DB. 


Exs.  410-422.]  SIMILAR   FIGURES.  193 


EXERCISES. 

410.    Prove  the  converse  of  prop.  XXII  :  If  the  perpendicular  drawn 
from  the  vertex  of    a  triangle  to  the  base  is  the  mean  proportional 
i       between  the  segments  of  the  base,  the  triangle  is  right-angled. 

Jfflr\   411.    Prove   that  any   chord  of    a  circle   is  the   mean  proportional 
I    between  its  projection  on  the  diameter  from  one  of  its  extremities,  and 
the  diameter  itself. 

412.  In  the  figure  on  p.  192,  if  AD  represents  three  units,  and  DB 
represents  one  unit,  what  number  is  represented  by  CD  ? 

413.  Prove  that  if  a  perpendicular  is  let  fall  from  any  point  on  a  cir- 
cumference, to  any  diameter,  it  is  the  mean  proportional  between  the 
segments  into  which  it  divides  that  diameter. 

414.  Prove  that  if  two  fixed  parallel  tangents  are  cut  by  a  variable 
IV     tangent,  the  rectangle  of  the  segments  of  the  latter  is  constant. 

415.  Through  any  point  in  the  common  chord  of  two  intersecting 
j£.    circumferences  two  chords  are  drawn,  one  in  each  circle.     Prove  that 

the  four  extremities  of  these  chords  are  concyclic. 

416.  If  the  bisectors  of  the  interior  and  exterior  angles  at  B,  in  the 
figure  of  prop.  XXII,  meet  b  at  F  and  E,  respectively,  prove  that  BC  is 
the  mean  proportional  between  FC  and  CE. 

i/       417.    Calculate  each  of  the  segments  into  which  the  bisectors  of  the 
y\  angles  of  a  triangle  divide  the  opposite  sides,  the  lengths  of  the  sides 

being  9  in.,  12  in.,  and  15  in.,  respectively. 
J&       418.    From  the  points  A,  B,  on  a  line  AB,  25  in.  long,  perpendiculars 

AC,  BD  are  erected  such  that  AC  =  13  in.,  BD  =  7  in.     On  AB  the 

point  0  is  taken  such  that  ZBOD  =  ZCOA.     Calculate  the  distances 

AO,  OB. 

419.  Given  a  trapezoid  ABCD,  with  the  non-parallel  sides  AD,  BC 
divided  at  E,  F,  respectively,  in  the  ratio  of  2  to  3,  to  calculate  the  length 
of  EF,  knowing  that  AB  =  12.45  in.,  and  DC  =  38.5  in. 

420.  Calculate  the  sides  of  a  right  triangle,  knowing  that  their  respec- 
tive projections  on  the  hypotenuse  are  2.88  in.  and  5.12  in. 

421.  The  two  sides  of  a  right  triangle  are  respectively  10  in.  and  24  in. 
Required  the  lengths  of  their  projections  on  the  hypotenuse,  and  the 
distance  of  the  vertex  of  the  right  angle  from  the  hypotenuse.    (To  0.001.) 

422.  The  two  sides  of  a  right  triangle  are  respectively  3.128  in.  and 
4.275  in.  Required  the  lengths  of  the  two  segments  into  which  the 
bisector  of  the  right  angle  divides  the  hypotenuse.     (To  0.001.) 


194  PLANE    GEOMETRY.  [Bk.  IV. 

6.     PROBLEMS. 

Proposition  XXIII. 

271.    Problem.     To  divide  a  line-segment  into  parts  propor- 
tional to  the  segments  of  a  given  line. 


Given         the  line  OX',  and  the  line  OX  divided  into  segments 
OA,  AB, 

Required    to  divide   OX'  into  segments  proportional  to   OA, 
AB,  

Construction.    1.  Placing  the  lines  oblique  to  each  other  at  a 
common  end-point  0,  draw  XX' .  §  28 

2.  From  A,  B,  draw  lines  II  XX',  cutting   OX'  at 

A',  B',  I,  prop.  XXXIII 

Then  OX'  is  divided  as  required. 

Proof.         '.'  OX,  OX'  are  two  transversals  of  a  pencil  of  ll's,  the 
corresponding  segments  are  in  proportion.      Prop.  X 

Corollaries.  1.  A  given  line  can  be  divided  into  parts 
proportional  to  any  number  of  given  lines. 

For  that  number  of  given  lines  may  be  laid  off  as  OA,  AB,  BC,  

on  OX. 

2.  A  line  can  be  divided  into  any  number  of  equal  parts. 

Note.  While  a  straight  line  can  be  divided  into  any  number  of  equal 
parts,  by  means  of  the  straight  edge  and  the  compasses,  a  circumference 
cannot  be  divided  into  7,  9,  11,  13,  and,  in  general,  any  prime  number  of 
equal  parts  beyond  5.     The  exceptions  are  noted  in  Book  V. 


Prop.  XXIV.]  PROBLEMS.  .  1(J5 

Proposition  XXIV. 

272.    Problem.      To  find  the  fourth  proportional  to  three 
given   lines. 


Given         three  lines,  a,  b,  c. 

Required    to  find  x  such  that  a  :  b  =  c  :  x. 

Construction.  1.  From  the  vertex  of  a  pencil  of  two  lines,  with 
the  compasses  lay  off  a,  b,  in  order,  on  one  line,  and 
c  on  the  other  line. 

2.  Join  the  end-points  of  a,  c,  remote  from  the  vertex, 
by  I  §  28 

3.  From  the  end-point  of  b,  remote  from  a.  draw  a  line 
parallel  to  I.  I,  prop.  XXXIII 

This  will  cut  off  x,  the  line  required. 
Proof.  a  :  b  —  c  :  x.  Prop.  X,  cor.  1 

273.    Definition.     If  a  :  b  =  b  :  x,  x  is  called  the  third  pro- 
portional to  a  and  b. 

Corollary.      The  third  proportional  to  two  given  lines  van 
be  found. 

For  to  find  x  such  that  a  :  b  =  b  :  x.  make  c  =  b  in  the  above  solution. 


Exercises.  423.  The  problem  admits  of  a  considerable  variation  of 
the  figure,  as  suggested  by  the  figure  given  in  ex.  383.  Invent  another 
solution  from  this  suggestion. 

424.  How  many  inches  in  the  fourth  proportional  to  lines  respectively 
2  in..  3  in..  5  in.  long?  In  the  third  proportional  to  lines  respectively 
2  in. ,  7  in.  long  ? 


196 


PLANE    GEOMETRY. 


[Bk.  IV. 


Proposition  XXV. 

274.    Problem.      To  find    the   mean  proportional    between 
two  given 


0       D       B 


Given         two  lines,  AD,  DB. 

Required    to  find  the  mean  proportional  between  them. 

Construction.    1.  Placing  AD,  DB  end  to  end  in  the  same  line, 
bisect  AB  at  0.  I,  prop.  XXXI 

2.  With  center  0  and  radius  OB,  describe  a  circle. 

§  109 

3.  Prom  D  draw  DC  J- AB,  to  meet  circumference 
at  C.  I,  prop.  XXIX 
Then  CD  is  the  mean  proportional. 

Proof.         AD  :  CD  =  CD  :  DB.    Prop.  XXII,  cor.  2,  and  §  238 

275.  Definition.  A  line  is  said  to  be  divided  in  extreme  and 
mean  ratio  by  a  point  when  one  of  the  segments  is  the  mean 
proportional  between  the  whole  line  and  the  other  segment. 

Thus,  AB  is  divided  internally  in  extreme  and  mean  ratio  at  P,  if 
AB  :  AP  =  AP  :  PB  ;  and  externally  in        f                                     p 
such  ratio  at  P',  if  AB  :  AP*  =  AP' :  P'B.      E 1 

To  say  that  AB  :  AP  =  AP  :  PB  is  A  B 

merely  to  say  that  AP2  =  AB  •  PB.     This  division  is  often  known  as 
the  Golden  Section  or  the  Median  Section. 

If  the  student  understands  quadratic  equations  he  will  see  that  if  the 
length  of  AB  is  6,  and  if  AP  =  x,  then  PB  =  G  —  x,  and 
v  AP2  =  AB  •  PB, 
...  x2  =  6  (6  -  x), 
or  x2  +  6x  -  36  -  0.     Solving,  x  =  -  3  =fc  3  VE. 


Prop.  XXVI.] 


PROBLEMS. 


197 


Proposition  XXVI. 
276.    Problem.      To  divide  a  line  i)i  extreme  and  mean  ratio. 


Given 


the  line  AB. 


Required    to  divide  AB  in  extreme  and   mean    ratio ;   i.e.  to 
find  P  such  that  AB  •  PB  =  AP2. 

Construction.    1.  Draw  CB  J_  AB  and  —\AB. 

2.  Describe  a  O  with  center  C  and  radius  CB. 

3.  Draw  AC  cutting  the  circumference  in  A' and  Y. 

4.  Describe  two  arcs  with  center  A  and  radii  AX  and 
A  Y,  thus  fixing  points  P,  P'. 

These  are  the  required  points. 
Proof  for  point  P.  Proof  for  point  P'. 


AB2  =  AX-  AY 

=  AP(AX+XY) 
=  AP  (AP  +  AB) 
=  AP2  +  AP  •  AB. 
.' .  AB  (AB  -  AP)  =  AP2. 
.'.AB'PB  =  AP2. 


AB2  =  AY-  AX 

=  P'A  (A  Y  -  XY) 
=  P'A  (P'A  -  AB) 
=  P'A2  -  AB  •  P'A. 

AB  (AB  +  P'^)  =  P'A2. 
.'.AB-P'B  =  P'A2. 


.'.  AB  is  divided  internally  at  P  and  externally  at  P'  in 
Golden  Section. 

It  should  be  noticed  that  if  the  sense  of  the  lines  as  positive  or  nega- 
tive is  considered  (that  is,  considering  AP  —  —  PA),  the  above  solutions 
would  be  identical  if  X  and  Y  were  interchanged,  and  P'  substituted 
for  P. 


198 


PLANE    GEOMETRY 


[Bk.  IV. 


Proposition  XXVII. 

277.  Problem.  On  a  given  line-segment  as  a  side  corre- 
sponding to  a  given  side  of  a  given  polygon,  to  construct  a 
polygon  similar  to  that  polygon. 

D      ^C 


Fig.  2. 

Given         the  polygon  ABCD  and  the  line-segment  A'B'. 

Required    to  construct  on  A'B'  as  a  side  corresponding  to  A B, 
a  polygon  A'B' CD'  ~  ABCD. 

Construction.    1.  In  Fig.  1,  place  A'B'  II  AB.    I,  prop.  XXXIII 

Draw  AA',  BB',  meeting  at  0;  draw  OC,  OD.    §  28 

Draw  B'C  II  BC,   CD'  II  CD.  I,  prop.  XXXIII 

Draw  D'A'.     Then  A'B'C'D'  —  ABCD. 

Proof.    1.  v  OA:OA'=OB:OB'=OC:OC'=OD:OD',  §244 

.-.  D'A'  II  DA.  Prop.  XII 

2.  v  A'B':AB=  OB':  OB  =  B'C:BC  = » 

Prop.  X,  cor.  3 
and  Z  C'B'A'  =  Z  CBA,  and  so  for  the  other  A, 

I,  prop.  XVII,  cor.  2;  ax.  3 
.  • .  A'B'  CD'  ~  AB  CD.  Prop.  XXI 

If  A'B'  =  AB,  as  in  Fig.  2,  draw  from  C,  D  IPs  to  AA'; 
otherwise  the  construction  is  as  above.  It  is  left  to 
the  student  to  prove  D'A'  II  DA,  and  A'B'C'D'  ^ 
ABCD.     .'.  A'B'C'D'  ~~  ABCD  by  §  262,  cor.  1. 


BOOK    V.  —  MENSURATION    OF   PLANE    FIGURES. 
REGULAR   POLYGONS   AND    THE    CIRCLE. 


1.    MENSURATION   OF  PLANE   FIGURES. 

Proposition  I. 

278.    Theorem.      Tivo  rectangles  having  equal  altitudes  are 
proportional  to  their  bases.  , 


Given         two  rectangles  R  and  R',  with  altitude  a,  and  with 
bases  b,  b',  respectively. 

To  prove    that  R  :  R'  =  b:b'. 

Proof.    1.  Suppose  b  and    b'  divided  into  equal  segments,  /, 
and  suppose  b  =  nl,  and  b'  =  n'l. 

(In  the  figures,  n  =  6,  n'  =  4.) 
Then  if  _k  are  erected  from  the  points  of  division, 

R  =  n  congruent  rectangles  al, 
and    R'  =  ?i'         "  "  « 


R 
R' 


n  -  al 

u'  -al 


Why  ? 


Note.  The  above  proof  assumes  that  b  and  b'  are  commensurable, 
and  hence  that  they  can  be  divided  into  equal  segments  I.  The  proposi- 
tion is,  however,  entirely  general.  The  proof  on  p.  200  is  valid  if  b  and 
b'  are  incommensurable. 

199 


200  PLANE  GEOMETRY. 

279.    Proof  for  incommensurable  case. 


R' 


[Bk.  V. 


■ 


1.  Suppose  b  divided  into  equal  segments  I, 
and  suppose  b  =  nl, 

while  b'  =  ?i'l  +  some  remainder  x, 

such  that  x  <  I. 

Then  if  _L's  are  erected  from  the  points  of  division, 
R  =  n  congruent  rectangles  al, 


and  22'=  ? 

such  that 


ax  <  al. 


al  +  a  remainder  ax, 
Why  ? 


2.  Then  b'  lies  between  n'l  and  (V  +  1)  I, 
(In  the  figure,  between  4 1  and  5  Z. ) 

and  B'  lies  between  n'  •  <zZ  and  (V  +  1)  ■  al.     Why  ? 


// 


7." 


3.  .*.  —  and  —  both  lie  between  —  and 
b  R  n 

(In  the  figure,  between  \  and  f.) 
and  .*.  they  differ  by  less  than  -• 
(In  the  figure,  by  less  than  \.) 


w'  +  l 


Why  ? 
Why? 


4.  And  v  -  can  be  made  smaller  than  any  assumed 

difference,  by  increasing  n, 

.'.to  assume  any  difference  leads  to  an  absurdity. 

K  V     R'      ,  R      b 

5.  .'.-  =  — ?  whence —=-• 

b       R  R'      b' 


Note.     The  proof  will  be  noticed  to  be  essentially  that  of  pp.  171, 181. 


Prop.  II.l      MENSURATION   OF  PLANE   FIGURES. 


201 


Corollaries.     1.  Rectangles  having  equal  bases  are  pro- 
portioned  to  their  altitudes. 

For  they  can  be  turned  through  90°  so  as  to  interchange  base  and 
altitude. 

2.  Triangles  having  equal  altitudes  are  proportional  to  their 

bases;  having  equal  bases,  to  their  altitudes.     (Why?) 

3.  Parallelograms   having  equal  bases   are  proportional  to 

their  altitudes;  having  equal  altitudes,  to  their  bases. 


Proposition  II. 

280.  Theorem.  Two  rectangles  have  the  same  ratio  as  the 
products  of  {the  numerical  measures  of)  their  bases  and 
altitudes. 


Given         two  rectangles  E,  Pi,  with  bases  b,  b',  and  altitudes 
a,  a\  respectively. 

To  prove    that  B:B'=  ab  :  a'b'. 

Proof.    1.  Let  X  be  a  rectangle  of  altitude  a  and  base  b'. 
E      b 

X_^o_ 
E'~  a'' 

2.  .-.—  =  — — ,  by  multiplying  corresponding 

members  of  the  two  equations.  Ax.  6 

Note.  Thus  again  appears  the  relation  between  geometry  and  algebra 
set  forth  in  §  221,  that  to  the  product  of  two  numbers  corresponds  the 
rectangle  of  two  lines. 


Then 
and 


Prop.  I 
Prop.  I,  cor.  1 


202  PLANE    GEOMETRY.  [Bk.  V. 

281.  Definition.  To  measure  a  surface  is  to  find  its  ratio  to 
some  unit.  The  unit  of  measure,  multiplied  by  this  ratio,  is 
called  the  area. 

Thus,  in  a  surface  4  ft.  long  by  2  ft.  broad,  the  ratio  of  the  surface  to 
1  sq.  ft.  is  8,  and  8  sq.  ft.  is  the  area. 

Corollaries.  1.  Parallelograms  (or  triangles)  have  the  same 
ratio  as  the  products  of  their  bases  and  altitudes.     (Why  ?) 

For  a  parallelogram  equals  a  rectangle  of  the  same  base  and  the  same 
altitude,  II,  prop.  I,  cor.  1.    See  also  (for  the  triangle)  II,  prop.  II,  cor.  1. 

2.  The  area  of  a  rectangle  equals  the  product  of  its  base  and 
altitude. 

That  is,  the  number  which  represents  its  square  units  of  area  is  the 
product  of  the  two  numbers  which  represent  its  base  and  altitude. 

For  in  prop.  II,  if  B'  =  1,  the  square  unit  of  area,  then  a'  and  b'  must 
each  equal  1,  the  unit  of  length.     Hence  R/l  =  ab/l,  or  E  =  ab. 

3.  The  area  of  a,  parallelogram  equals  the  product  of  its  base 
and  altitude  ;  of  a  triangle,  half  that  product. 

See  the  proof  under  cor.  1. 

4.  The  area  of  a  square  equals  the  second  power  of  its  side. 
This  is  the  reason  that  the  second  poiver  of  a  number  is  called  its  square. 

5.  The  area  of  a  trapezoid  equals  the  product  of  its  altitude 
and  half  the  sum  of  its  bases.     (Why  ?) 

See  II,  prop.  III. 

Exercises.  425.  Prove  that  any  quadrilateral  is  divided  by  its  inte- 
rior diagonals  into  four  triangles  which  form  a  proportion. 

426.  ABC  is  a  triangle,  and  P  is  any  point  in  BC ;  from  P  are  drawn 
two  parallels  to  CA,  BA,  meeting  AB,  AC  in  X,  Y,  respectively.  Prove 
that  A  AXY  is  a  mean  proportional  between  A  BPX  and  A  PCY.  In- 
vestigate when  P  is  on  CB  produced. 

427.  Suppose  D,  E,  the  mid-points  of  sides  b,  a  of  A  ABC,  to  be 
joined  ;  draw  AE  and  ED,  intersecting  at  0.  Prove  that  ABEO  is  a 
mean  proportional  between  A  DOE  and  ABO.  Investigate  when  BE  II 
AB,  but  D  and  E  are  not  mid-points  of  b,  a. 


Prop.  III.]     MENSURATION   OF  PLANE   FIGURES.  203 

Proposition  III. 

282.  Theorem.  Triangles,  or  parallelograms,  which  have 
an  angle  in  one  equal  to  an  angle  in  the  other,  have  the  same 
ratio  as  the  products  of  the  including  sides. 


A  B      B 

Given         two  triangles  ABC,  AB'C,  having  an  angle,  A,  of 
one  equal  to  an  angle,  A,  of  the  other. 

To  prove    that  A  ABC  :  A  AB'C  =  AB  ■  AC  :  AB'  •  AC. 

Proof.    1.   Suppose  them  placed  with  Z A  in  common;  draw 
BC,  B'C 
Then  AABCiA  AB'C  =  AB  :  AB'.  Why  ? 

2.  And  AAB'C:AAB'C  =  AC:  AC, 

their  bases  being  AC,  AC.  Why  ? 

3.  .-.  A  ABC:  A  AB'C  =  AB  -  AC  :  AB' ■  AC.       Ax.  6 

4.  And  '•'  the  UJ  in  the  figure  are  double  the  A,  the 
theorem  is  true  for  parallelograms. 

Corollary.  Similar  triangles  have  the  same  ratio  as  the 
squares  of  their  corresponding  sides. 

For  if  the  A  are  similar,  BC  II  B'C.  and  the  ratio  AB  :  AB'  equals, 
and  may  be  substituted  for,  the  ratio  AC  :AC\  thus  making  the  second 
member  of  step  3,  AB2  :  AB'2. 


Exercises.  428.  Prove  prop.  Ill,  changed  to  read,  "an  angle  in  one 
supplemental  to  an  angle  in  the  other." 

429.  Prove  the  converse  of  prop.  Ill,  cor.  :  If  two  triangles  have  the 
same  ratio  as  the  squares  of  any  two  corresponding  sides,  they  are 
similar. 


204  PLANE    GEOMETRY.  [Bk.  V. 

Proposition  IV. 

283.    Theorem.     Similar  polygons  have  the  same  ratio  as 
the  squares  of  their  corresponding  sides. 


Given         P  and  P',  two  similar  polygons;  sides  a,  b,  , 

corresponding  to  sides  a',  b', ;  diagonal  s  corre- 
sponding to  diagonal  s'. 

To  prove    that  P  :  P'  =  a2 :  a'2. 

Proof.  1.  Suppose  P  and  P'  divided  into  similar  A  of  bases 
a  and  a',  b  and  b',  ,  by  diagonals  from  corre- 
sponding points  0,  0'.  IV,  prop.  XXI,  cor.  4 

Then  A  Oa  :  A  O'a'  =  a2 :  a'2,  Prop.  Ill,  cor. 

and     A  Oa  :  A  O'a'  =  s2 :  s'2  =  A  Ob  :  A  O'b'  = 

Why? 

2.  .'.AOft  +  OH :  A  O'a'  +  O'b'  + =  AOa:A  O'a'. 

IV,  prop.  VI 

3.  '.'AOa  +  Ob  + =  P,  and  A  0'a'+  0'b'+ =  P', 

.'.P:P'  =  a2:a'2. 


Exercises.     430.    If  the  vertices,  A,  B,  C,  of  a  triangle  are  joined  to 
a  point  0  within  the  triangle,  and  if  AO  produced  cuts  a  at  D,  then 
A  ABO:  A  AOC  =  BD:  DC. 

431.  If  two  triangles  are  on  equal  bases  and  between  the  same  paral- 
lels, then  any  line  parallel  to  their  bases,  cutting  the  triangles,  will  cut 
off  equal  triangles. 

432.  Two  equilateral  triangles  have  their  areas  in  the  ratio  of  1  : 2. 
Find  the  ratio  of  their  sides  to  the  nearest  0.01. 


Props.  V,  VI.]     PARTITION   OF   THE  PERIGON.  205 

2.     PARTITION   OF   THE   PERIGON. 

Proposition  V. 
284.    Problem.      To  bisect  a  perigon. 

Construction  and  Proof.    A  special  case  under  I,  prop.  XXVIII. 
Corollary.     A  perigon  com  be  divided  into  2n  equal  angles. 


Proposition  VI. 
285.    Problem.      To  trisect  a  perigon. 

R 


Given         the  perigon  with  vertex  0. 

Required    to  trisect  it. 

Construction.    1.  On  any  line  OA,  from  0,  construct  an  equi- 
lateral A  OAB.  Authority  ? 

2.  Produce  AO  to  C,  and  bisect  Z  COB  by  OB. 
Then  the  perigon  is  trisected  by  OB,  0  C,  OB. 
Proof.    1.  v  ZAOB  =  60°,  I,  prop.  XIX,  cor.  8 

.-.ZBOC,  supplement  of  Z.AOB  =  120°. 

2.  .'.  Z  COB,  conjugate  of  Z  BOC     =  240°. 

3.  .-.  Z  COD  =  Z BOB  =  120°.  Const.  2 


206 


PLANE    GEOMETRY. 


[Bk.  V. 


Corollary.  A  perigon  can  be  divided  into  3  •  2"  equal 
angles. 

For  if  n  =  0,  then  3  •  2M  =  3  •  1  =  3,  so  that  the  corollary  reduces  to  the 
problem  itself.  If  n  =  1,  then  3  •  2n  =  6,  and  by  bisecting  A  BOC,  COD, 
DOB,  the  perigon  is  divided  into  6  equal  angles.  Similarly,  by  bisecting 
again,  the  perigon  is  divided  into  3  •  22  =  12  equal  angles,  and  so  on. 


Proposition  VII. 
286.     Problem.     To  divide  a  perigon  into  five  equal  angles. 
Y 


Given 


0     M     P  A 

the  perigon  with  vertex  0. 


Required    to  divide  it  into  five  equal  angles. 

Construction.    1.  Draw  OA,  and  divide  it  at  P  so  that  OP  •  OA 
=  PA2.  IV,  prop.  XXVI 

2.  Draw  MY,  the  _L  bisector  of  OP.        I,  prop.  XXXI 

3.  With  center  P  and  radius  PA  describe  an  arc  cutting 
MYinB.  §109 


4. 

Draw  OB. 

Then 

Z  A  OB  =  £  of  a  perigon. 

Proof.    1. 

Draw  AB  and  PB. 

Then 

v  OP-  OA  =  PA2, 

and 

v  OA  >  PA, 

.' 

.  OP  <  PA,  and  .'.MP  <  PA. 

.-.incuts  MY. 

Prop.  VII.]  PARTITION    OF    THE   PEIUGON.  207 

2.  v  A  OPB  is  isosceles,  OB  =  PB  =  PJ,  the  radius. 

3.  Also,  0Aa  +  OB1  =  AB2  +  2  0Jf  •  OA 

II,  prop.  IX,  cor.  1 
And  •.-2  03I=OP, 

.'.  041  +  OP2  =  ^P2  +  OP  ■  OA. 
4.  And      V  OB2  =  PA2  =  OP  •  OJ, 

.-.  OA2  +  OP2  =  ^4P2  +  OB2,  from  step  3, 
.-.  OA2  =  AB2, 
and  .-.  0.1  =  AB. 

5.  .-.  Z  0P.4  =  Z0  =  Z  BPO,  I,  prop.  Ill 

=  AA  +  Z.  PBA.       I,  prop.  XIX 
And  v  £A  =  £  PBA,  I,  prop.  Ill 

.\Z0  =  2ZJ. 

6.  .'.  Z  0  is  f  of  a  st.  Z,  or  £  of  a  perigon. 

I,  prop.  XIX 

Corollary,    yl  perigon  can  be  divided  into  5  •  ,J"  equal  parts. 

For  if  n  =  0,  then  5  •  2n  =  5  •  1  =  5,  so  that  the  corollary  reduces  to  the 
problem  itself.  If  n  =  1,  then  5  •  2n  =  5  •  2  =  10,  and  by  bisecting  Z  A  OB, 
the  resulting  angle  is  jL  0f  a  perigon.  Similarly,  by  bisecting  again,  ^  0f 
a  perigon  is  formed,  and  so  on. 


Exercises.  433.  In  the  figure  on  p.  200,  let  OP=  x,  PA  =  r ;  then  show 
that  x--(VE  —  1).  (Omit  exs.  433,  434  if  the  student  has  not  had 
quadratic  equations.) 

434.  In  the  same  figure,  if  OP  =  x  and  OA  =  a,  show  that  x  = 
|(S-V6). 

435.  On  the  sides  a,  b,  c.  of  an  equilateral  triangle,  points  X,  Y,  Z 
are  so  taken  that  BX  :  XG  -  CY :  YA  =  AZ  :  ZB  =  2  : 1.  Find  the 
ratio  of  A  XYZ  to  A  .4£C. 


208 


PLANE   GEOMETRY. 


[Bk.  V. 


Proposition  VIII. 
287.    Problem.     To  divide  a  perigon  into  fifteen  equal  angles. 

D\      \C     /B 


A 

Solution.  1.  Make       Z.  AOB  —  \  of  a  perigon.        Prop.  VII 

2.  Make  Z  ^40D  =  ^  of  a  perigon.  Prop.  VI 

3.  Bisect  Z  BOD.  I,  prop.  XXVIII 

4.  Then  /.  BOC '=£(£  —  £)  perigon  =  ^  of  a  perigon. 

Corollary.     ^4  perigon  can  be  divided  into  15  •  #*  equal 
angles. 
Explain. 

288.  Note.  That  a  perigon  could  be  divided  into  2",  3  ■  2",  5  ■  2W, 
15  •  2n  equal  angles,  was  known  as  early  as  Euclid's  time.  By  the  use  of 
the  compasses  and  straight  edge  no  other  partitions  were  deemed  possible. 
In  1796  Gauss  found,  and  published  the  fact  in  1801,  that  a  perigon  could 
be  divided  into  17,  and  hence  into  17  •  2n  equal  angles  ;  furthermore,  that 
it  could  be  divided  into  2m  -f  1  equal  angles  if  2m  +  1  was  a  prime  number ; 
and,  in  general,  that  it  could  be  divided  into  a  number  of  equal  angles 
represented  by  the  product  of  different  prime  numbers  of  the  form 
2m  -f  1.  Hence  it  follows  that  a  perigon  can  be  divided  into  a  number 
of  equal  angles  represented  by  the  product  of  2n  and  one  or  more  different 
prime  numbers  of  the  form  2m  -f  1.  It  is  shown  in  the  Theory  of  Num- 
bers that  if  2m  +  1  is  prime,  m  must  equal  2p  ;  hence  the  general  form  for 
the  prime  numbers  mentioned  is  22P  +  1.  Gauss's  proof  is  only  semi- 
geometric,  and  is  not  adapted  to  elementary  geometry. 


Exercises.     436.    Including  the  divisions  of  a  perigon  suggested  by 
Gauss,  there  are  25  possible  divisions  below  100.     What  are  they  ? 

437.    As  in  ex.  43G,  there  are  13  possible  divisions  between  100  and 
300.     What  are  they  ? 


Prop.  IX.]  REGULAR  POLYGONS.  209 

3.     REGULAR   POLYGONS. 

Proposition  IX. 

289.    Problem.      To  inscribe  in  a  circle  a  regular  'polygon 
having  a  given  number  of  sides. 


Given         a  circle  with  center  0  and  radius  OA. 

Required    to  inscribe  in  the  circle  a  regular  ?i-gon. 

Construction.  1.  Divide  the  perigon  0  into  n  equal  parts  (n 
being  limited  as  in  props.  V-VIII  and  cors.)  as  AOB, 
BOC,  COD, ,  B,  C,  D, lying  on  the  circum- 
ference. 

2.  Draw  AB,  BC,  CD §  28 

Then  ABCD is  an  inscribed  regular  %-gon. 

Proof.    1.        AAOB^ABOC^ACOD^ , 

and      AB  =  BC  =  CD  =  I,  prop.  I 

2.  .'.  AB  =  BC=  CD= Ill,  prop.  IV 

3.  .\ZDCB  =  Z.  CBA  = , 

'.'  each  staDds  on  (n  —  2)  arcs  equal  to  AB. 

Ill,  prop.  XI,  cor.  1 

4.  .".  ABCD is  an  inscribed  regular  polygon. 

§§  92,  201 


210 


PLANE   GEOMETRY. 


[Bk.  V. 


Corollaries.     1.    The  side  of  an  inscribed  regular  hexagon 
equals  the  radius  of  the  circle. 

Then  Z  AOB  =  \  of   360°  =  GO0 ;   .-.  Z  BA  0,  which  =  Z  OB  A  =  60°. 
.-.  A  ABO  is  equilateral. 

2.  An  inscribed  equilateral  polygon  is  regular. 

For  by  step  3  of  the  proof  it  is  also  equiangular ;  and  being  both  equi- 
lateral and  equiangular,  it  is  regular. 


Proposition  X. 

290.    Problem.      To  circumscribe  about  a  circle  a  regular 
polygon  having  a  given  number  of 


Given         a  circle  with  center  0  and  radius  OA. 

Required    to  circumscribe  about  this  circle  a  regular  ?i-gon. 

Construction.  1.  Divide  the  perigon  O  into  n  equal  parts  {ii 
being  limited  as  in  props.  V— VIII  and  cors.)  by  lines 
OW,  OX,  OY, 

2.  Bisect  A  WOX}  XO  Y, by  radii  to  A,  Ii, 

I,  prop.  XXVIII 

3.  From  A,  B,  C, draw  tangents  to  meet  OJJ'at  7>, 

OX  at  E, Ill,  prop.  XXVI 

Then  DEFG is  the  required  polygon. 


Prop.  X.] 


REG  ULA  R   POL YG OXS. 


211 


Proof.    1.  v  DE±  OA,  EF±  OB,  ,      III,  prop.  IX,  cor.  2 

.-.A  OAE^A  OBE,  and  AE  =  BE,  OE  =  OK 

I,  prop.  II 

2.  .'.  the  tangents  from  A  and  7>  meet  0 X  at  the  same 
point,  E. 

3.  And  v  Z.DOE  =  A  EOF, 
and  Z  OEZ)  =  Z  i^O, 
.'.  A  DOE^A  EOF,  and  Z>#  =  ## 


Const.  1 
Step  1 
Why  ? 


4.  Also,  v  Z  GEE  =  A  FED,  each  being  the  supple- 
ment of  an  Z  equal  to  Z  WOX,     (Xame  the  Z.) 

I,  prop.  XXI,  cor. 

.'.  DEFG is  a  circumscribed  regular  polygon. 

§§  92,  201 

Corollaries.     1.    The  side  of  a  regular  hexagon  circum- 
scribed about  a  circle  of  diameter  1,  is  1/V3,  or  i  v3. 

For  it  is  (as  in  prop.  IX,  cor.  1)  the  side  of  an  equilateral  A  whose 
altitude  is  ^.     This  is  easily  shown  to  be  1/Vs.     (Show  it.) 

2.  A  circumscribed  equiangular  polygon  is  regular. 
Prove  that  any  two  adjacent  sides  are  equal. 


Exercises.     438.    In  a  right-angled  triangle,  any  polygon  on  the  hypote- 
nuse equals  the  sum  of  two  similar  polygons  described  on  the  sides  as 
corresponding  sides  of  those  polygons. 
(Suggestion  :    P2  :  Pz  =z  b2  :  c2  ; 

.-.  P2  +  P3:P3  =  &a  +  c2  :  C2 

=  a2  :  c2  =  P1:PZ; 

.-.  P2  +  P3  :  Pi  =  P3  :  Ps  =  1.     This  is  one    s 
of  the  generalized  forms  of  the   Pythago- 
rean theorem.) 

439.  If  r  is  the  radius  of  the  circle,  and 
8  is  the  side  of  the  inscribed  equilateral 
triangle,  then  s  =  r  V3. 


212  PLANE   GEOMETRY.  [Bk.  V. 

Proposition  XI. 

291.    Problem.     To    circumscribe    a    circle    about   a   given 
regular  polygon. 

E       D 

C 

B 


Given         the  regular  polygon  A  BCD 

Required    to  circumscribe  a  circle  about  it. 

Construction.    Bisect  A  DCB,  CBA,  the  bisectors  meeting  at  0. 
Then  0  is  the  center  and  OB  the  radius. 

Proof.    1.  Draw  OA,  OD,  OE, Then  V  A  OCB,  CBO  are 

halves  of  oblique  A,  each  is  less  than  a  rt.  Z. 

2.  .'.  CO  and  BO  cannot  be  II,  and  they  meet  as  at  0. 

3.  And       v  ZCBO  =  Z  OB  A,  Const, 
and                   AB  =  BC,               §  92,  clef.  reg.  pol. 

4.  .'.A  ABO  £  A  C50,  and  6U  =  OC.  Why  ? 
Similarly  each  of  the  lines  OB,  OD, =  OC. 

5.  .*.  0  is  the  center,  and  OA,  OB, are  radii.  §  108 

292.  Note.  The  inscription  and  circumscription  of  regular  polygons 
are  seen  to  depend  upon  the  partition  of  the  perigon.  Elementary  geometry 
is  thus  limited  to  the  inscription  and  circumscription  of  regular  polygons 
of  2",  3-2",  5  •  2",  15  •  2n  sides  ;  or,  since  the  discovery  by  Gauss,  to  poly- 
gons the  number  of  whose  sides  is  represented  by  the  product  of  2n  and 
one  or  more  different  prime  numbers  of  the  form  2m  +  1. 

In  addition  to  regular  convex  polygons,  cross  polygons  can  also  be 
regular,  the  common  five-pointed  star  being  an  example. 


Prop.  XII.] 


REGULAR   POLYGONS. 


213 


Proposition  XII. 

293.    Problem.     To  inscribe  a    circle    in  a   given  regular 
polygon. 


Given         a  regular  polygon  WXY 

Required    to  inscribe  a  circle  in  it. 

Construction.    1.  Circumscribe  a  circle  about  it.  Prop.  XI 

2.  From  center  0  of  tins  O  draw  OA  _L  WX. 

I,  prop.  XXX 
With  center  0,  and  radius  OA,  a  O  may  be  inscribed. 

Proof.    1.  Draw  OB,  OC,  _L  XY,  YZ, 

Then  v  OA  bisects  WX,  .'.A  lies  between  W  and 
X,  and  so  for  B,  C, Ill,  prop.  V 

2.  And*.'  WX=XY= ,  .\  OA  =  OB  = 

Ill,  prop.  VII 

3.  .'.if  with  center  0  and  radius  OA  a  O  is  described, 
then  WX,  XY, will  be  tangent  to  the  O, 

III,  prop.  IX,  cor.  3 
and  .'.  the  O  is  inscribed  in  the  polygon. 

§  201,  def.  inscr.  O 


Exercises.  440.  Solve  prop.  XI  by  bisecting  the  sides  AB,  BC  by 
perpendiculars,  thus  determining  0. 

441.  Inscribe  a  regular  cross  pentagon  in  a  circle.  (The  regular  cross 
pentagon,  the  pentagram,  was  the  badge  of  the  Pythagorean  school.) 


214  PLANE    GEOMETRY.  [Bk.  V. 

Corollaries.  1.  The  inscribed  and  circumscribed  circles  of 
a  regular  polygon  are  concentric. 

For  from  step  2  of  the  construction  and  step  2  o^f:  the  proof,  0  is  the 
center  of  both  circles. 

2.  The  bisectors  of  the  angles  of  a  regular  polygon  meet  in 
the  common  in-  and  circumcenter. 

For  by  the  proof  of  prop.  XI  they  meet  in  O,  and  by  cor.  1  0  is  the 
common  in-  and  circumcenter. 

3.  The  perpendicular  bisectors  of  the  sides  of  a  regular 
polygon  meet  in  the  common  in-  and  circumcenter.     (Why  ?) 

294.  Definitions.  The  radius  of  the  circumscribed  circle  is 
called  the  radius  of  a  regular  polygon  ;  the  radius  of  the 
inscribed  circle,  the  apothem  of  that  polygon ;  the  common 
center  of  the  two  circles,  the  center  of  that  polygon. 

E.g.  in  the  figure  below,  r  is  the  radius,  m  the  apothem,  and  0  the 
center  of  the  regular  polygon,  part  of  which  is  shown  as  inscribed  in  the 
circle. 


Proposition  XIII. 

295.    Theorem.     The  area  of  a  regular  polygon  equals  half 
the  product  of  the  apothem  and  perimeter. 


Given         an  inscribed  regular  polygon,  of  area  a,  perimeter 
p,  apothem  m. 

To  prove    that  a  =  ^  mp. 


Prop.  XIII.]  REGULAR   POLYGONS.  215 

Proof.         Let  0  be  the  center  and  /•  the  radius  of  the  circum- 
scribed circle. 

Let  t  be  one  of  the  A  formed  by  joining  0  to  two 
consecutive  vertices,  and  s  a  side  of  the  polygon. 
Then  area  t  equals  -J-  ms.  "Why  ? 

.'.  the  area  of  the  polygon  equals  the  sum  of  the 
areas  of  the  triangles  =  \m  X  the  sum  of  the  sides 
=  \  mp.  Ax.  2 

Corollaries.  1.  The  areas  of  regular  polygons  of  the 
same  number  of  sides  are  proportional  to  the  squares  of  their 
%pothemSj  of  their  radii,  or  of  their  sides. 

For  -  =  i^L  =  J^L  ;    and  from    similar  &  and   IV,  prop.  XX, 
a       $ mp       mp 

m       r       s       p         .         .    .j.  a       m2       r2       s2 

—  =  —  =  —  =  —•  .-.by  substitution  —  =  —^  =  —  =  —  • 
m       r       s       p  a       m'2      r2      s- 

2.  The  'perimeters  of  regular  polygons  of  the  same  number 
of  sides  are  proportional  to  their  apothems,  their  radii,  or 
their  sides. 

Proved  with  cor.  1. 

Exercises.  442.  The  distance  from  the  center  to  a  side  of  the  inscribed 
equilateral  triangle  equals  r/2. 

443.  Draw  a  diameter  AB  of  a  circle  with  center  0 ;  then  with 
center  A  and  radius  A  0  draw  an  arc  cutting  the  circumference  in  C,  D  ; 
draw  CD.  DB,  BC,  and  prove  A  BCD  equilateral. 

444.  The  area  of  an  inscribed  regular  hexagon  is  a  mean  propor- 
tional between  the  areas  of  the  inscribed  and  circumscribed  equilateral 
triangles. 

445.  Show  how,  with  compasses  alone,  to  divide  a  circumference  into 
six  equal  arcs. 

446.  Prove  that  if  AB,  CD,  two  diameters  of  a  circle,  are  perpen- 
dicular to  each  other,  then  ACBD  is  an  inscribed  square. 

447.  Let  OX  be  the  perpendicular  bisector  of  line-segment  AB  at  0  ; 
lay  off  on  OX,  OD  =  AO  ;  and,  on  DX,  lay  off  DC  =  DB  ;  then  prove 
that  C  is  the  center  of  the  O  circumscribed  about  the  regular  octagon  of 
which  AB  is  a  side. 


216 


PLANE   GEOMETRY. 


[Bk.  V. 


4.     THE  MENSURATION  OF  THE  CIRCLE. 

296.  Postulate  of  Limits.  The  circle  and  its  circumference 
are  the  respective  limits  which  the  inscribed  and  circum- 
scribed regular  polygons  and  their  perimeters  approach,  if 
the  number  of  their  sides  increases  indefinitely. 


The  following  may  be  read  by  the  student  in  connection 
with  the  postulate,  although  it  does  not  constitute  a  proof : 

1.  In  the  figure,  suppose  an  in-  and  circumscribed  regular  ?i-gon  rep- 

360° 
resented.    Then  each  exterior  angle  equals in  each  figure. 

360°  180° 

2.  .-.  each  interior  angle  equals  180° >  and  .-.  Z  a  =  90° 

°  n  n 


3.  .-.  if  n  increases  indefinitely,  Za  =  90°,  and p  =  r. 

4.  .-.  the  inscribed  polygon  =  the  circle,  and  its  perimeter 
circumference.     Similarly  for  the  circumscribed  polygon. 


the 


Coeoll aries.  1.  The  circumscribed  regular  polygon  and 
its  perimeter  are  respectively  greater  than  the  circle  and  its 
circumference  ;  the  inscribed,  and  its  perimeter,  less. 

2.  If,  on  any  finite  closed  curve,  n  points 
are  assumed  equidistant  from  each  other,  and 
each  connected  with  the  succeeding  point  by 
a  straight  line,  then  the  curve  is  the  limit  which  the  broken 
line  approaches  if  n  increases  indefinitely. 


Prop.  XIV.]       MENSURATION   OF   THE   CIRCLE.  217 

Proposition  XIV. 

297.  Theorem.  The  ratio  of  the  circumference  to  the 
diameter  of  a  circle  is  constant. 

Proof.  1.  Suppose  any  two  circles,  of  circumferences  c,  c', 
radii  r,  r',  and  diameters  d,  d',  respectively,  to  have 
similar  regular  polygons  inscribed  in  them,  of 
perimeters  p,  p\  respectively. 

Then  p  :p'  =  r  :  >•',  Prop.  XIII,  cor.  2 

=  2r:2r'  =  d:  d'.  IV,  prop.  VIII 

2.  And  v  >\  ?•',  d,  d'  do  not  change  when  the  number  of 
sides  of  the  polygons  is  doubled,  quadrupled, , 

§  294,  def.  radius  polyg. 

and  '.'  p  =  c,  and  p*  =  e',  §  296,  post,  of  limits 

r.c:c'  =  d:d'.  IV,  prop.  IX 

3.  .' .  c  :  d  =  c' :  d'  =  the  same  for  any  (D.  IV,  prop.  Ill 

Note.  This  constant  ratio  c  :  d  is  designated  by  the  symbol  it  (pi), 
the  initial  letter  of  the  Greek  word  for  circumference  (periphereia). 
The  value  of  it  is  discussed  in  prop.  XVII. 

Corollaries.     1.  c  =  7rd,  or  2tti\ 

For  if  -  =  7t,  then  c  =  nd. 
d 

2.  If  the  radius  of  a,  circle  is  1,  then  Q,  =  2ir,  or  a  semi- 
circumference  equals  it. 

3.  The  circumferences  of  two    circles    are  proportional  to 

their  radii. 

_,      c       2  nr       r 

For  -  = =  —  • 

c/      2  itr'      r' 

Exercises.    448.  Find,  in  terms  of  the  radius  of  the  circle,  r,  the  side, 
apothem,  and  area  of  the  inscribed  and  circumscribed  equilateral  triangle. 
449.    Also  of  the  inscribed  and  circumscribed  square. 


218 


PLANE    GEOMETRY. 


[Bk.  V. 


Proposition  XV. 


298.  Problem.  Griven  the  sides  of  the  regular  inscribed 
and  circumscribed  n-gons,  to  find  the  side  of  the  regidar 
circumscribed  2  n-gon. 


A'  M'      C 

Solution.     1.  In  the  figure, 

let  AB  =  a  side  of  the  regular  inscribed     w-gon,  in  ; 
"  A'B'  =       "  «  "  circumscribed       "       en. 

Then  BM ' 

=       "         "  "     inscribed     2?i-gon,  i2n; 

and  BC 

=  J  "         "  "  circumscribed       "       c2n. 

2.  But  v  OC  bisects  Z  M'OB',  Why  ? 

.'.  CB':M'C=  OB':  OM'  (=  OB), 

IV,  prop.  XIII 
=  A'B' :  .47?,  IV,  prop.  X,  cor.  3 
=  c«  ■  k> 

3.  .'.  CB'  +  M'C :  J/'C  =  cn  +  s  :  in,  IV,  prop.  V 
or              M'B':M'C  =  cn  +  in  :  /„. 

4.  . ' .        2  M'B'  :2M'C  =  cn  +  in  :  in.        IV,  prop.  VIII 

5.  .-.     ,  cB:c2B  =  cfI  +  /„:/,„ 


or 


Hn 


0n  +  '» 


Prop.  XVI.]       MENSURATION   OF    THE    CIRCLE.  219 

Proposition  XVI. 

299.  Problem.  Given  the  sides  of  the  regular  inscribed 
n-gon  and  the  regular  circumscribed  2  n-gon,  to  find  the  side 
of  the  regular  inscribed  2  n-gon. 

Solution.    1.   In  the  figure  on  p.  218, 

A  M'BM  -  A  CM'D.         .  Why  ? 

.\  M'B  :  BM  =  CM' :  JI'I).  Why  ? 

2-   Or                    h.  :*<.  =  **.:*»,,,  Why? 


* 


.•.^  =  iV2^./,,  Why? 


Corollaries.     1.  //'pn.  p2n>  Pn?  ^n  represent  the  perimeters 
of  the  polygons  with    sides   in,  i2n,  cn,  c2n,  respectively,   then, 

(!)  P^»  =  ¥irt'  and  (2)  ft-  =  viv^- 

rn     I     pn 

For  c2n  =  Pi>„/2?i,  cw  =  P„/w,  /*„  =p2«/-«,  and  i„  =  pn/n;  substi- 
tute these  in  the  tinal  steps  of  props.  XV,  XVI.     From  prop.  XV, 

P-j»  _    P„/n-p„/n    _  Pn-Pn/n 
2n       Pn/n  +  pn  /  »        P«  +  P» 


2"       P»+P» 


From  prop.  XVI, 

P2 

2ra 


»    =  I        /JT  5J?  .  Pn 

i       2  \       2n     n 


P2n=   ^Pn-P'2i 


2.  cn  =  —  =?  where  r  ?s  f/^p  radius. 


Vr2-iin2 


>2f  =  f  :  Jf2  -  (|  V  =  r  :  Vr^  -  i  tf. 
/'  l"  by  multiplying  by  in. 


220  PLANE   GEOMETRY.  [Bk.  V. 

Proposition  XVII. 

300.   Theorem.     The  approximate  value  of  it  is  3.14159  +. 

Proof.    1.  In  a  regular  hexagon  inscribed  in  a  circle  of  diam- 
eter 1,  iG  =  J,  and  .'.^6  =  3.  Prop.  IX,  cor.  1 

2.  Of  the  regular  hexagon  circumscribed  about  that  O, 
c6  =  l/V3.  Prop.  X,  cor.  1 

3.  .\Pa  =  6-c6  =  3.4641016 

4.  From  jh  and  P6  can  be  found  p^  and  P12. 

Props.  XV,  XVI 

5.  From  p>\2  and  P12  can  be  found  ^>24  and  P24,  and 
so  on.  Props.  XV,  XVI 
If  the  process  were  continued  to  a  1536-gon,  ^1536 
would  be  found  to  be  3.1415904,  and  P1536  would  be 
found  to  be  3.1415970. 

6.  And  '.'  c,  or  ird,  which  equals  it  •  1  or  ir,  lies  between 
pn  and  P„,  however  large  n  may  be, 

§  296,  post,  of  limits,  cor.  1 
.".  7T  lies  between  3.1415904  and  3.1415970,  and  is, 
'  therefore,  approximately  3.14159  +. 


Exercises.     450.    The  diagonals  of  a  regular  pentagon  cut  each  other 
in  extreme  and  mean  ratio. 

451.  If  ABODE  is  a  regular  pentagon,  and  AD  cuts  BE  at  P,  prove 
that  AP  :  AE  =  AE  :  AD. 

452.  To  construct  a  regular  pentagon  equal  to  the  sum  of  two  given 
regular  pentagons. 

453.  Find,  in  terms  of  the  radius  of  the  circle,  r,  the  side  of  the 
inscribed  regular  pentagon.     (Omit  unless  ex.  433  was  taken.) 

454.  Also  of  the  inscribed  and  circumscribed  regular  hexagon. 

455.  Also  of  the  inscribed  and  circumscribed  regular  dodecagon. 

456.  Also  of  the  inscribed  regular  decagon.     (Depends  on  ex.  453.) 


Secs.  301,  302.]       MENSURATION   OF    THE   CIRCLE.  221 

301.   Notes.     The  computation  in  prop.  XVII,  which  the 
student  is  not  expected  to  make,  is  as  follows : 


So.  of  sides 

6 

p 
3. 

p 

3.4641016 

12 

3.1058285 

3.2153903 

24 

48 

3.1326286 
3.1393502 

3.1596599 
3.1460862 

96 

3.1410319 

3.1427146 

192 

3.1414524 

3.1418730 

384 

768 

3.1415576 

3.1415838 

3.1416627 
3.1416101 

1536 

3.1415904 

3.1415970 

302.  The  following  historical  notes  on  it  are  inserted  to  show 
the  student  how  the  subject  of  the  mensuration  of  the  circle 
has  grown. 

The  early  approximation  for  7T,  in  use  among  the  ancient  people,  was  3. 
See  I  Kings,  vii,  23;  II  Chron.  iv,  2.  "What  is  three  hand-breadths 
around  is  one  hand-breadth  through. "  —  The  Talmud. 

Ahmes,  however,  gave  the  equivalent  of  3.1604. 

Archimedes  seems  to  have  been  the  first  to  employ  geometric  methods 
similar  to  that  of  props.  XV,  XVI  for  approximating  it.  He  announced, 
"The  circumference  of  a  circle  exceeds  3  times  the  diameter  by  a  part 
which  is  less  than  i,  but  more  than  i^,  of  the  diameter." 

Hero  of  Alexandria  used  both  3  and  3i. 

Ptolemy  of  Alexandria  gave  3Tyo- 

Aryabhatta  found  3.1416,  by  a  method  similar  to  that  of  prop.  XVII. 

Brahmagupta  used  the  values  of  Archimedes;  also  f§§£  and  ||ib  tne 
last  being  only  another  form  for  Ptolemy's. 

Metius  gave  the  easily  remembered  value  355/113. 

Ludolph  van  Ceulen  computed  it  to  the  equivalent  of  over  30  decimal 
places  (the  decimal  fraction  was  not  yet  invented),  and  wished  it  engraved 
on  his  tomb  at  Leyden.  On  this  account  it  is  often  called  in  Germany, 
"the  Ludolphian  number." 

Vega  carried  it  to  140  decimal  places. 

Dase  carried  it  to  200  decimal  places. 

Kichter  carried  it  to  500  decimal  places.  More  recently  Shanks  carried 
it  to  707  decimal  places. 

The  symbol  it  is  first  used  in  this  sense  in  Jones's  "  Synopsis  Palma- 
riorum  Matheseos,"  London,  1706. 


222  PLANE   GEOMETRY.  [Bk.  V. 

303.  Definition.  It  is  now  necessary  fro  extend  our  idea  of 
equal  surfaces.  The  definition  at  the  beginning  of  Book  II, 
§  142,  is  true,  and  it  suffices  for  the  cases  there  under  con- 
sideration. But  when  curvilinear  figures  are  compared  with 
rectilinear,  it  is  impossible  to  cut  the  surfaces  into  parts 
respectively  congruent.  Hence,  we  enlarge  the  definition, 
thus :  Two  surfaces  are  said  to  be  equal  if  they  have  the 
same  numerical  measure  in  terms  of  a  common  unit. 

Thus,  a  circle  having  an  area  of  2  m2  would  equal  a  rectangle  2  m  long 
by  1  m  broad,  even  though  they  could  not  be  cut  into  parts  respectively 
congruent. 

304.  Table  of  Values.  The  following  table  of  values  of 
expressions  involving  ir  will  be  found  useful  in  computations 
concerning  the  circle,  sphere,  cylinder,  cone,  etc. : 

it  =  3.14159  Vtt  =  1.77245  180°/ tt  =  57°.  29578 

;r/4  =  0.78540  1/Vtt  =  0.56419  tt/180  =    0.01745 

1  /  it  =  0. 31831  it  V2  =  4. 44288  Approximate  values : 

7T2  =  9.86960  V^72  =  1.25331                      %  =  -2T2-  =  3},  fff. 

The  table  is  repeated,  with  other  tables  of  value  in  numerical  compu- 
tations, at  the  end  of  this  work. 

305.  Radian  Measure  of  Angles  and  Arcs.  Since  if  A  =  any 
central  angle  and  a  =  its  arc, 

Aist.Z.        =  a:  semi#ircumf.  =  a  :  wr. 
.'.  A  :  st.  Z./ir  =  a:r, 
or  A:  180° /ir  =a:r, 
or  ^:57°.29+  =  a  :  r. 

That  is,  the  ratio  of  a  central  angle  to  st.  Z./nr  equals  the 
ratio  of  its  arc  to  an  arc  of  the  same  length  as  the  radius. 
Just  as  the  "degree"  is  the  unit  for  both  angle  and  arc 
measure,  it  being  understood  to  be  ^^  of  a  perigon  in  the 
one  case  and  ^^  of  a  circumference  in  the  other,  so  a  special 
name  is  given  to  st.  Z /ir  and  to  an  arc  which  equals  a  radius 
in  length  ;    this   name   is  radian.      In  other  words,  a  radian 


>  Y 

Sec.  306.]  MENSURATION   OF   THE   CIRCLE.  223 

is  —  of  a  st.  Z.,  in  angle  measure,  and  —  of  a  semicircumfer- 

IT  IT 

ence,  or  an  arc  equal  to  a  radius  in  length,  in  arc  measure. 
Since  r  =  —  180°,  .*.  r  =  57°.29  +  ,  where  r  stands  for  radian. 

IT 

irr  it 

Also  v  180°  =  7T/-,      .".  1°  =  — — -  or  — —  of    a  radian,   or 

.01745.33  of  a  radian. 

In  most  work  in  advanced  mathematics  the  radian  measure 
is  used  exclusively.  In  common  measurements  the  degree  is 
used.     It  is  necessary  in  this  work  to  use  both. 

It  is  customary  to  express  an  angle  in  radians  by  the  Greek 

letters  a  (alpha),  (3  (beta),  y  (gamma),  ,  the  first  letters  of 

that  alphabet. 

306.  Corollary.  The  length  of  an  arc  equals  the  product 
of  the  radius  by  the  angle  i In  radians. 

For  if  a  =  length  of  arc,  and  a  =  its  Z  in  radians,  then  -  =  — , 

9  ^v  C  2  7T 

.-.  a  =  a =  a  ■  r. 

27t  

Exercises.  457.  Express  the  following  in  radians  :  10°,  21°  20',  57°, 
58°,  90°. 

458.  Express  the  following  in  degrees:  1.3090 r,  .8058  r,  .3636  r, 
.1687  r,  .0029  r. 

459.  Express  the  following  in  radians  :  100°,  180°,  270°. 

460.  Express  the  following  in  degrees  :  3.4907  r,  5.2359  r,  0.2832  r,  nr. 

461.  Find  the  lengths  of  arcs  of  47°  50',  61°  20',  75°  40',  the  radius 
being  10. 

462.  Given  the  lengths  of  the  following  arcs,  to  find  the  radii  of  the 
various  circles  :    75°  10',  131.19  ;   32°  20',  2.822  ;   4°,  .0698. 

463.  Show  that  the  perimeters  of  the  inscribed  and  circumscribed 
squares,  the  diameter  of  the  circle  being  1,  are  respectively  2.8284271 
and  4 ;  hence,  find  the  perimeters  of  the  inscribed  and  circumscribed 
regular  octagons,  and  thus  show  that  the  value  of  it  may  be  approxi- 
mated in  this  way. 

464.  The  circumferences  of  certain  ©  are  43.9823,  84.8230,  128.8053, 
185.5340,  204.2035  ;   find  the  diameters. 


224  PLANE    GEOMETRY.  [Bk.  V. 

Proposition  XVIII. 

307.   Theorem.     The  area  of  a  circle  equals  half  the  prod- 
uct of  its  circumference  and  radius. 

Given         a,  c,  r,  the  area,  circumference,  and  radius  of  a  circle. 

To  prove    that  a  =  ^  cr. 

Proof.  1.  If  a',  p  represent  the  area  and  perimeter  of  a  cir- 
cumscribed regular  polygon,  then  the  apothem  of 
that  polygon  is  r.  §  294 

2.  And     a'  =  \pr.  Prop.  XIII 

3.  But      a'  =  a,  and  \  pr  =  \  cr.     §  296,  post,  of  limits 

4.  .'.  a  =  \ cr.  IV,  prop.  IX,  cor.  1 

Corollaries.     1.  a  =  ttv2. 
For  c  =  2  7tr. 

2-«  =  £-    (Why?)   . 

3.  If  s  represents  the  area  of  a  sector,  and  a  its  angle  in 
radians,  then  s  =  v2a/2. 

For  s  :  itr2  =  a  :  2  it.     (IV,  prop.  XVI,  cor.) 

4.  Of  two  unequal  circles,  the  greater  has  the  greater  circum- 
ference. 

c2 

For,  by  cor.  2,  a  = 

4  it 

.-.  4  na  =  c2. 
.-.  as  the  area  increases,  the  circumference  increases  also. 

5.  The  areas  of  two  circles  are  proportional  to  the  squares 
of  their  radii. 

„      a       itr2       r2 

For  -  =  — -  =  — 

a       itr2      r"2 


Sec.  308.]  MENSURATION   OF   THE   CIRCLE.  225 

308.  Historical  Note  on  Quadrature  of  the  Circle.  The  ex- 
pression, "  to  square  the  circle,"  means  to  find  the  side  of  a  square  whose 
area  equals  that  of  a  given  circle.  The  solution  of  this  problem  by 
elementary  geometry  has  been  proved  to  be  impossible.  It  nevertheless 
occupied  the  attention  of  many  mathematicians  before  this  impossibility 
was  shown,  and  many  ignorant  people  still  attempt  it.  Some  of  the 
Pythagorean  school  claimed  to  have  solved  it,  Anaxagoras  (died  428  b.c.) 
wrote  upon  it,  and  hundreds  of  writers  since  then  have  discussed  the  sub- 
ject. It  is  closely  related  to  finding  a  straight  line  equal  to  a  given  circum- 
ference ("to  rectify  the  circumference"),  and  the  two  depend  upon 
finding  the  value  of  it  exactly.  That  it  cannot  be  expressed  exactly,  nor 
as  the  root  of  a  rational  algebraic  equation,  was  shown  by  Lindemann 
in  1882. 

For  the  mathematical  discussion,  see  Klein's  "Famous  Problems  of 
Elementary  Geometry,"  translated  by  the  authors.    (Boston,  Ginn  &  Co.) 


Exercises.     465.    What  is  the  radius  of  that  circle  of  which  the  number 
of  square  units  of  area  equals  the  number  of  linear  units  of  circumference  ? 

466.  Also,  of  which  the  number  of  square  units  of  area  equals  the 
number  of  linear  units  of  radius  ? 

467.  Give  a  formula  for  a  in  terms  of  d,  and  the  constant  it. 

468.  A  circle  equals  a  triangle  of  which  the  base  equals  the  circumfer- 
ence and  the  altitude  equals  the  radius. 

469.  Find  the  areas  of  circles  with  radii  5,  7,  21,  35,  47,  50.    (In  these 
computations,  for  uniformity  let  it  =  3.1416.)  j 

470.  Also  with  diameters  2,  8,  11,  31,  42,  97.  ^ 

471.  Find  the  radii  of  circles  of  areas  78.5398,  2042.8206,  4536.4598. 

472.  Also  the  diameters  of  circles  of  areas   2123.7166,  3318.3072, 
56.745017. 

473.  Also  the  circumferences  of  circles  of  areas  95.0332,  452.3893. 
\y       474.    Also  the  areas  of  circles  of  circumferences  267.0354,  191.6372. 

475.    The  area  of  the  ring  formed  between  the  circumferences  of  two 


* 


concentric  circles  of  radii  rj.,  r2,  where  T\  >  r2,  is  it  (rx  +  r2)  (j\  —  r2). 

476.  The  area  of  that  portion  of  the  ring  of  ex.  475  cut  off  by  the  arms 
of  the  central  angle  a  radians  is  \a{rx  i-  r2)  (ri  —  r2) ;  or,  if  ai,  a2  are 
arcs  bounding  that  portion,  the  area  =  I  (cii  +  a2)  (ri  —  r2). 


Note.    The  remainder  of  the  work  may  be  omitted  without  destroying 
the  integrity  of  the  course. 


APPENDIX   TO   PLANE   GEOMETRY. 


1.      SUPPLEMENTARY  THEOREMS   IN   MENSURATION. 

Proposition  XIX. 

309.    Theorem.     If  the  sides  of  a  triangle  are  a,  b,  c,  and 
if  s  =  i  (a  +  b  4-  c),  sx  =  s  —  a,  s2  =  s  —  b,  s3  =  s  — <3,  then  the 


area  equals  V s  •  81  -  s2  •  s3. 


Fig. 


b 
Fig.  2. 


Proof.    1.  a2  =  b2  +  c2^2  be',  2  be'  taking  the  sign  -  for  Fig.  1, 
+  for  Fig.  3,  and  being  0  for  Fig.  2.  §  159 

.*.  c'  =  ±  (b2  +  c2  —  a2)  /2  b,    by   solving  the  above 
equation  for  c'.  Axs.  2,  7 

2.  But  h2  =  c2  -  c'2  =  (c  +  c')  (c  -  c')  §  154 
=  [e  +  (b2  +  c2  -  a2)  /2  ft]  [c  -  (62  +  c2-  a2)  /2  ft],  by 
substituting  the  value  of  c'  given  in  step  1. 

.-.  K1  =  (2  be  +  £2  +  c2  -  a2)  (2  6c  -  62  -  c2  +  a2)  /4  ft2, 
by  removing  parentheses  and  simplifying. 

3.  .'.  4  b2h2  =  [<7,  +  c)2  -  a2]  [a2  -  (b  -  c)2],  by  multi- 
plying by  4  &2  and  factoring. 

.-.4 ^A2  =  (b  +  c -f  </)(/,  +  c  - a)(a  +b-  c)(a  -b  +  c), 
by  factoring  still  farther. 
226 


Prop.  XX.]  THEOREMS   IX  MENSURATION. 


227 


4.  But  if  a  +  b  +  c  =  2  s.  as  given, 
then  6  +  c  —  a  =  2  (s  -  a)  =  2  sb 

and    a  —  b  +  c  =  2 (s  —  b)  =  2s2, 
and    a  +6  -  c  =  2(s  -  c)  =  2s3. 

5.  .•.4&2A2  =  2s.2s1-2s2-2s3.  Subst.  in  3 

6.  .'.  area  =  \bh  —  Vs  •  sx  ■  s2  ■  s3.        Y.  prop.  II,  cor.  3 

Note.  This  is  known  as  Hero's  formula  for  the  area  of  a  triangle.  Of 
course  a,  b,  c  represent  numerical  values  as  explained  under  V,  prop.  II, 
cor.  2. 


Proposition  XX. 

310.    Theorem.      The  radius,  r,  of  the  circle  circumscribed 
about  the  triangle  abc  of  area  t,  equals  abc/4t. 

c 


Proof.    1.   Suppose  CX  =  d  a  diameter,  CD  (or  A)  _L  AB.  and 
2LY  drawn. 

Then  A  ADC  ~  A  X£C,  IV,  prop.  XYII 

and  .\d:a  =  b:h.  Why? 

2.  .-./•  =  a£/2  A-        IV,  prop.  I ;   ax.  7 

3.  But  v  i  he  =  t.  V,  prop.  II,  cor.  3 

.-.  r=  abc  jit.  Subst.  3  in  2 

Note.     The  value  of  t  can  be  found  by  Hero's  formula. 


Exercise.     477.    Find  the  areas  of  the  triangles  with  sides  (1)  13,  14. 
15;  (2)  3,  o,  8  ;   (3)  7,  10,  18;   (4)  a,  a,  a;  (5)  3,  4,  5. 


228  PLANE   GEOMETRY.  [Bk.  V. 

Proposition  XXI. 

311.  Theorem.  The  product  of  the  diagonals  of  an  inscrip- 
tible  quadrilateral  equals  the  sum  of  the  products  of.  the 
opposite  sides. 

D 


Given  ABCD,  an  inscriptible  quadrilateral,  with  sides  a,  b, 
c,  d,  and  diagonals  e,  f. 

To  prove    that  ef=ac-\-  bd. 

Proof.  1.  Let  ABCD  be  inscribed,  the  sides  arranged  as  in  the 
figure,  chord  AK  =  BC,  and  DK  drawn  cutting  AC 
at  L. 

Then  Z  ADK  =  Z  BDC,  and  Z  CAD  =  Z  CJ5D, 
and  .-.  AALD^  ABCD.  Why? 

2.  Also  v  Z  DG'X  =  Z  Di?.4,  and  Z  LDC  =  Z  .4D^, 

III,  prop.  XI,  cor.  1 ;   ax.  2 
.-.  A  CDL  ~  A  BDA.  Why  ? 

3.  From  1,  AL  :d  =  b:e,  or  ^4Z  =  bd/e;    IV,  prop.  XX 
from   2,  LC  :c  =  a:e,  or  LC  =  ac/e.      IV,  prop.  XX 

4.  .'.  ^LL  +  LC,  or  JC,  or/=  (ac  +  M)/e.  Ax.  2 

5.  .'.  ef=  ac  +  6d.  Ax.  6 

Note.     Ptolemy's  theorem. 


Exercise.     478.    Is  prop.  XXI  true  when  b  =  zero  ? 


(ty 


Secs.  312,  313.]  MAXIMA    A  XL   MI XI MA.  229 


2.     MAXIMA   AND   MINIMA. 

312.  Definitions.  If  a  geometric  magnitude  can,  by  con- 
tinuous change,  increase  until  a  value  is  reached  at  which  the 
magnitude  begins  to  decrease,  such  value  is  called  a  maximum 
value ;  if  it  can  similarly  decrease  until  a  value  is  reached  at 
which  it  begins  to  increase,  such  value  is  called  a  minimum 
value. 

In  general,  a  magnitude  can  have  more  than  one  maximum  or 
minimum  value,  as  in  the  annexed  /""~rX 

figure  where   «i,    «2,   a$   represent        _  s^\  / 

maximum,    and    &i,    &g,    minimum        /A  ^^r'a  i  ^^r  a3' 

values  of  the  ordinates  of  F.     In    ^        '•      "■!  !      ^i [_^ 

the  elementary  geometry  of  the  line 

and  circle,  however,  only  one  maximum  or  minimum  exists,  so  that  the 

words  here  mean  greatest  and  least. 

E.g.  the  maximum  chord  of  a  circle  is  the  diameter  (III,  prop.  VIII, 
cor.),  and  the  minimum  chord  is  spoken  of  as  zero,  since  zero  is  the  limit 
which  constantly  decreasing  chords  of  a  circle  approach. 

A  magnitude  at  its  maximum  value  is  called  a  maximum;  similarly, 
a  minimum.  E.g.  a  chord  of  a  circle  is  a  maximum  when  it  is  a 
diameter. 

313.  Figures  having  equal  perimeters  are  said  to  be  isoperi- 
metric. 


Exercises.  479.  Draw  a  line  AB,  bisect  it  at  M,  and  take  a  point 
X  on  AM ;  then  show  that  AX1  +  XB2  =  2  AM-  +  2  XM2,  and  that 
this  is  a  minimum  when  XM  =  0 ;  hence  show  that  the  sum  of  the 
squares  on  the  two  segments  of  a  given  line  is  a  minimum  when  the  seg- 
ments are  equal. 

480.  Also  that  AX  •  XB  =  MB2  —  XM'2,  and  that  this  is  a  maximum 
when  XM  =  0  ;  that  is,  that  the  rectangle  of  the  two  segments  into  which 
a  given  line  can  be  divided  is  a  maximum  when  the  given  line  is  bisected. 

481.  If  the  diagonals  of  an  inscribed  quadrilateral  are  perpendicular 
to  each  other,  then  the  sum  of  the  products  of  the  two  opposite  sides 
equals  twice  the  area  of  the  quadrilateral. 


230  PLANE    GEOMETRY.  [Bk.  V. 

Proposition  XXII. 

314.  Theorem.  Of  all  triangles  formed  with  the  same  two 
given  sides,  that  is  the  maximum  ivhose  sides  contain  a  right 
angle. 


AD  B 


Given         the  A  AB  Cl9  AB  C2,  with  A  Cx  =  A  C2,  and  A  C2  _L  AB. 
To  prove    that  A  ABC2  >  A  ABCV 

Proof.         Suppose  C\D  J_  AB. 

Then  AC\  >  DCl9  I,  prop.  XX 

and  .'.  its  equal  AC2  >  BC\. 

.'.AABC2>  AABC19      II,  prop.  II,  cor. 3 

since  they  have  the  same  bases  but  different  altitudes. 


Exercises.  482.  Find  in  radians  the  angle  a:  of  a  sector  of  a  circle  of 
radius  r,  such  that  the  number  of  square  units  of  its  area  equals  the 
number  of  linear  units  of  its  entire  perimeter. 

483.  Interpret  the  result  of  ex.  482  for  r  =  2  ( 1  +  -  Y  Discuss  it  for 
r  <£  2.     Discuss  it  for  r  <  2  (l  +  -Y 

484.  In  the  Sulvasutras,  early  semi-theological  writings  of  the  Hindus, 
it  is  said:  "Divide  the  diameter  into  15  parts  and  take  away  2;  the 
remainder  is  approximately  the  side  of  the  square  equal  to  the  circle." 
From  this  compute  their  value  of  it. 

485.  On  AB  describe  a  semicircle,  and  in  it  inscribe  the  isosceles  tri- 
angle ABC ;  on  BC  and  CA  describe  semicircles  opposite  the  A  ABC. 
Show  that  A  ABC  =  the  sum  of  the  two  limes  thus  formed.  (The  limes 
of  Hippocrates.) 

486.  Six  lights  are  placed  regularly  on  the  circumference  of  a  circle 
of  radius  21  ft.;  what  are  the  distances  of  each  from  each  of  the  others  ? 
(To  0.01.) 


Prop.  XXIII.] 


MAXIMA    AX1)   MIX  IMA. 


231 


Proposition  XXIII. 

315.    Theorem.      Of  all  isoperimetric  triangles  on  the  same 
base  the  isosceles  is  the  maximum. 


^B' 


Given         two  isoperiuietric  A  ABC  and  ABX,  A  ABC  being 
isosceles,  with  AC  —  BC. 

To  prove    that  AABC>  A  ABX. 

Proof.    1.   On  AC  produced,  let    CB'  =  AC;   draw  B'B,  B'X; 
suppose  CD  II  AB. 
Then        v  AC  =  CB', 

.-.  BD  =  DB'.  I,  prop.  XXVII,  cor.  2 

2.  And         v  CB  =  AC, 

.-.  CB=  CB',  and  CDA.BB'.  Why  ?  Ax.  1 

3.  .  • .  A  C  +  CJB  =  ^  C  +  CT'  <  AX  +  X£'. 

Ax.  2 ;  I,  prop.  VIII 

4.  * . '  .IX  -f-  X£  =  A  C  +  G#,  Why  ? 
.'.  AT  +  X£  <  AX  +  X#', 

and  .-.  XB<XB'.  Why? 

5.  .'.  X  and  .IZ>  lie  on  the  same  side  of  CD, 

I,  prop.  XX,  cor.  3 
and  .'.A  ABC  >  A  ABX.  II,  prop.  II,  cor.  3 

Corollary.      Of  all  isoperimetric  triangles,  that  which  is 
equilateral  is  the  maximum.      (Why  ?) 


232  PLANE   GEOMETRY.  [Bk.  V. 

Proposition  XXIV. 

316.   Theorem.      Of  all  triangles  having  the  same  base  and 
area,  the  isosceles  has  the  minimum  perimeter. 


Given         the  A  AB  C  and  ABX  having  the  same  base  and 
area,  with  AC  =  BC. 

To  prove    that  perimeter  ABC  <  perimeter  ABX. 

Proof.    1.  Suppose  CYWAB;  AC  produced  so  that  CB'  =  AC; 
B'X  drawn ;  and  B'B  drawn  cutting  C  Y  at  D. 
Then    v  A  ABC  =  A  ABX, 

.'.  CY  passes  through  X.         II,  prop.  II,  cor.  4 

2.  And  v  AC=  CB', 

,\BD  =  DB'.  I,  prop.  XXVII,  cor.  2 

3.  And      v  A  BBC  ^  A  B'DC,  I,  prop.  XII 

.-.CDA.BB',  Why? 

and              .'.  XB  =  XB'.  I,  prop.  XX 

4.  But      AC  +  CB'  <  AX+  XB',  I,  prop.  VIII 
and  .'.  AC  +  CB  <  AX  +  XB. 

5.  .-.  perim.  ABC  <  perim.  ABX.  Why  ? 

Corollary.  Of  all  equal  triangles,  that  which  is  equi- 
lateral has  the  minimum  perimeter. 

For  whatever  side  is  taken  as  the  base,  the  perimeter  is  less  if  the  other 
two  sides  are  equal. 


Prop.  XXV.]  MAXIMA   AND  MINIMA.  233 

Proposition  XXV. 

317.  Theorem.  If  the  ends  of  a  line  of  given  length  are 
joined  by  a  straight  line,  and  the  area  of  the  figure  enclosed 
is  a  maximum,  it  takes  the  form  of  a  semicircle. 


P  __ 


Given         a   line  APB   (the   curve   in   the   figure),   of  given 
length,  and  AB  joining  its  end-points. 

To  prove    that,  if  the  area  of  the  figure  ABP  is  a  maximum, 
ABP  is  a  semicircle. 

Proof.  1.  Let  P  be  any  point  on  the  line;  then  joining  A  and 
P,  B  and  P,  let  the  segments  cut  off  by  AP,  BP  be 
called  s1}  s2,  and  A  ABP  called  t,  as  in  the  figure. 
Then  Z  P  is  a  right  angle ;  for  if  not,  without 
changing  su  s2,  the  area  of  t  could  be  increased 
by  making  Z  P  right.  Prop.  XXII 

2.  But  this  is  impossible  if  ABP  is  a  maximum,  and 
similarly  for  any  other  point  on  APB.  Why  ? 

3.  .'.  the  area  enclosed  is  a  maximum  when  the  line 
connecting  A  and  B  subtends  a  right  angle  at  every 
point  on  the  curve. 

Note.  It  will  be  seen  that  examples  of  maxima  or  minima  involve 
also  the  idea  of  symmetry  (§  68).  This  fact  is  of  value  in  solving  problems 
in  maxima  and  minima. 


Exercise.  487.  Given  the  points  A,  B,  on  the  same  side  of  line  X'X, 
to  find  on  X'X  a  point  P  such  that  Z  X'PA  =  Z  BPX.  Prove  that 
AP  +  PB  is  the  shortest  path  from  A  to  X'X  and  back  to  B.  (Reflected 
ray  of  light.) 


234 


PLANE   GEOMETRY. 


[Bk.  V. 


Proposition  XXYI. 

318.    Theorem.      Of   all    isoperimetric    plane    figures    the 
maximum  is  a  circle. 


Proof. 


Suppose  A,  B  points  bisecting  the  given  perimeter, 
AB  cutting  the  figure  into  two  segments,  su  s2. 


Then 


are  maxima  when  they  are  semicircles, 


and  AB  is  a  diameter, 


Why  ? 


Proposition  XXVII. 


319.    Theorem.      Of  all  equal  plane  figures  the  circle  has 
the  minimum  perimeter. 


Given         circle  C  —  plane  figure  P. 

To  prove    that  circumference  C  <  perimeter  P. 

Proof.    1.  Suppose  X  a  circle  of  circumference  equal  to  perim- 
eter P. 

Then  P  <  X,  Prop.  XXVI 

and  .'.  C  <  X.  Subst. 

2.  .*.  circumference  C  <  circumference  A',    §  307,  cor.  4 

and  .  * .  circumference  C  <  perimeter  P.  Subst. 


Prop.  XXVIII. 


MAXIMA    AND   MINIMA. 


235 


Proposition  XXVIII. 

320.    Theorem.     A  polygon  with  given  sides  is  a  maximum 
when  it  is  inscriptible. 


Given         two  polygons,  F  and  P',  with  given  sides  a,  b,  c , 

F  being  inscribed  in  a  circle,  and  F'  not  inscriptible. 

To  prove    that  F  >  F'. 

Proof.    1.  Name  the  circular  segments  on  a,  b, (opposite  P), 

A,  B, ;  suppose  congruent  segments  constructed 

on  a,  b,  (opposite  P'). 

Then  F  +  A  +  B  + >  F'  +  A  +  B  + 

Prop.  XXVI 


2. 


.\P>  P' 


Why? 


Exercises.  488.  If  the  diagonals  of  a  parallelogram  are  given,  its 
area  is  a  maximum  when  it  is  a  rhombus. 

489.  What  is  the  minimum  line  from  a  given  point  to  a  given  line  ? 
Where  has  this  been  proved  ? 

490.  Into  what  two  parts  must  a  given  number  be  divided  so  that  the 
product  of  those  parts  shall  be  a  maximum  ?     (Compare  ex.  479.) 

491.  As  a  corollary  to  ex.  479,  show  that  of  isoperimetric  rectangles 
the  square  is  the  maximum. 

492.  Find  the  point  in  a  given  straight  line  such  that  the  tangents 
drawn  from  it  to  a  given  circle  contain  the  maximum  angle. 

493.  A  straight  ruler,  1  foot  long,  slips  between  the  two  edges  of  the 
floor  (the  edges  making  a  right  angle).  Find  the  position  of  the  ruler 
when  the  triangle  formed  by  the  edges  and  ruler  is  a  maximum;  also 
the  area  of  that  triangle. 


236  PLANE    GEOMETRY.  [Bk.  V. 

Proposition  XXIX. 

321.    Theorem.      Of  all  isojjerimetric  polygons  of  a  given 
number  of  sides,  the  maximum  is  regular. 

B    X 


Given  P,  the  maximum  polygon  of  a  given  perimeter  and 
a  given  number  of  sides. 

To  prove    that  P  is  regular. 

Proof.  1.  Any  two  adjacent  sides,  AB,  BC,  must  be  equal. 
For  if  unequal,  as  AX,  XC,  then  A  AXC  could  be 
replaced  by  A  ABC,  having  AB  =  BC,  thus  enlarging 
P  without  changing  the  perimeter.  But  this  is  im- 
possible because  P  is  a  maximum.  Prop.  XXIII 

2.  And  hence  P  is  inscriptible  because  its  sides  are 
given.  Prop.  XXVIII 

3.  .".  P  is  regular.  V,  prop.  IX,  cor.  2 


Exercises.  494.  Considering  only  the  relation  of  space  enclosed  to 
amount  of  wall,  what  would  be  the  most  economical  form  for  the  ground 
plan  of  a  house  ? 

495.  Of  all  triangles  in  a  given  circle,  what  is  the  shape  of  the  one 
having  the  greatest  area  ?     Prove  it. 

496.  Through  a  point  of  intersection  of  two  circumferences  draw  the 
maximum  line  terminated  by  the  two  circumferences. 

497.  Of  all  triangles  of  a  given  base  and  area,  the  isosceles  has  the 
greatest  vertical  angle. 

498.  Draw  the  minimum  straight  line  between  two  non-intersecting 
circumferences. 


Prop.  XXX.]  MAXIMA    AXD   MINIMA.  237 

Proposition  XXX. 

322.    Theorem.      Of  two    isoperimetric    regular  polygons, 
that  having  the  greater  number  of  sides  is  the  greater. 


Proof.    1.  Let  ABCD  be  a  square,  P  a  point  on  DA,  A  PCX 
isoperimetric  with  A  PCD  and  having  CX=  PX. 
Then  A  PCX  >  A  PCD,  Prop.  XXIII 

and  .-.  pentagon  ABCXP  >  □  ABCD.  Ax.  4 

2.  But  pentagon  ABCXP  would,  with  the  same  perim- 
eter, be  greater  if  it  were  regular.  Prop.  XXIX 

3.  .'.  a  regular  pentagon  is  greater  than  an  isoperi- 
metric square.  Similarly,  a  regular  hexagon  would 
be  greater  than  an  isoperimetric  regular  pentagon, 
and  so  on. 

Exercises.  499.  A  cross-section  of  a  bee's  cell  is  a  regular  hexagon. 
Show  that  this  is  the  best  form  for  securing  the  greatest  capacity  with 
a  given  amount  of  wax  (perimeter). 

500.  Find  the  maximum  rectangle  inscribed  in  a  given  semicircle. 

501.  Find  the  minimum  square  inscribed  in  a  given  square. 

502.  Draw  the  minimum  tangent  from  a  variable  point  in  a  given 
line  to  a  given  circle. 

503.  What  is  the  area  of  the  largest  triangle  that  can  be  inscribed  in 
a  circle  of  radius  5  ? 

504.  Given  a  square  of  area  1.  Find  the  area  of  an  isoperimetric 
(1)  equilateral  triangle,   (2)  regular  hexagon,  (3)  circle. 


238 


PLANE    GEOMETRY. 


[Bk.  V. 


3.     CONCURRENCE   AND   COLLINEARITY. 
Proposition  XXXI. 

323.  Theorem.  If  X,  Y,  Z  are  three  points  on  the  sides  a- 
b,  c,  respectively,  of  a  triangle  ABC,  such  that  the  pterpen- 
diculars  to  the  sides  at  these  points  are  concurrent,  then 

(BX2  -  XC2)  +  (CY2  -  YA2)  +  (AZ2  -  ZB2)  =  0  ; 

and  conversely. 


Proof. 


Let  P  be  the  point  of  concurrence,  and  draw  PA, 
PB,  PC. 

Then  (BX2  -  XC2)  +  (CY2  -  YA2)  +  (AZ2  -  ZB2) 
=  PB2  -  PC2  +  PC2  -  PA2  +  PA2  -  PB2  =  0, 
for  BX2  -  XC2  =  (BP2  -  PX2)  -  (PC2  -  PX2)  = 
BP2  -PC2,  and  so  for  the  rest. 

Conversely:   1.  Suppose  the  _b  from  X,  Y,  to  meet  at  P; 
and  suppose  PZ'  _L  c. 
Then  as  above, 
(BX2  -  XC2)  +  (CY2-  YA2)  +  (AZ'2  -  ZB2)  =  0. 

2.  But  (BX2  -  XC2)  +  (CY2-  YA2)  +  (AZ2  -  ZB2)  =  0, 
and  AZ'2  -  Z'B2  =  AZ2  -  ZB2.  Why  ? 

3.  .'.  AZ'2  -  AZ2  =  Z'B2  -  ZB2-  but  those  differences 
have  opposite  signs  and  cannot  be  equal  unless  each 
is  zero. 

4.  .*.  Z  must  coincide  with  Z'. 


Prop.  XXXII.]    CONCURRENCE  AND    COLLINEARITY.         239 

Proposition  XXXII. 

324.  Theorem.  If  three  lines,  x,  y,  z,  drawn  from  the 
vertices  of  triangle  ABC  to  meet  a,  b,  c  in  X,  Y,  Z,  are  con- 
current, then 


AZ    EX    CI' 


ZB    XC 


— —  =  1 ;  and  conversely. 


Proof.  1.  Let  P  be  the  point  of  concurrence.  Then  v  A  APC, 
PBC  have  the  base  PC,  they  are  proportional  to 
their  altitudes,  and  .'.to  AZ,  ZB.  Why  ? 

AZ      A  .IPC 


2. 

and 
and 


ZP  A  PPG' 
BXABPA 
XC~  A  APC9 
CYAPBC 

YA~  ABBA 


_  .    AZ  BX  CY 

3-  '     ZBXC'YA  =  1-  Ax'6 

Conversely  :   Let  CP  meet  c  in  Z' ;   then  as  above, 
.4Z'  PA'  CT_ 
ZP  *  XC '  YA  ~ 

K  _  .  ^  px  cr 

°-But  zpxg'T^  =  l  Glven 


6. 


'  '  Z?B~  ZB 

7.  .'.  Z'  must  coincide  with  Z.  IV,  prop.  XI,  cor. 

Note.     Ceva's  theorem. 


240 


PLANE    GEOMETRY. 


[Bk.  V. 


Proposition  XXXIII. 

325.   Theorem.     If  three  points,  X,  Y,  Z,  lying  respectively 
on  the  three  sides  a,  b,  c  of  triangle  ABC,  are  collinear,  then 


and  conversely. 


AZ  BX  CY 
ZB ' XC ' YA 


-1 


2.  And  similarly, 

and 
3. 


Ax.  6 


Proof.    1.  Let  I,  m,  n  be  perpendiculars  from  A,  B,  C  onlF. 
Then  by  similar  A,  ZB  being  here  negative, 

AZ        I 
ZB~ -m 

BX  _m 
XC~ n 
CY  _n 
YA~ I 

AZ  BX  CY 
*''  ZB'  XC'  YA 

Conversely  :  Let  XY  meet  AB  in  Z' ;  then  as  above, 

AZ'   BX  CY 
Z'B' XC    YA 

AZ  BX  CY 
ZB  XC    YA 

AZ'     AZ 
•'*  Z'B~ ZB 

7.  .'.  Z'  must  coincide  with  Z.  IV,  prop.  XI,  cor. 

Note.     Menelaus's  theorem. 


4. 


But 


1. 


-1. 


Given 


Prop.  XXXIV.]     CONCURRENCE   AND   COLLINEARITT.        241 

Proposition  XXXIV. 

326.  Theorem.  If  a  circumference  intersects  the  sides, 
a,  b,  c,  of  a  triangle  ABC,  in  the  points  Ai  and  A2,  Bi  and  B2, 
Ci  and  C2,  respectively,  then 

ACi    BAi    CBi    AC2    BA2    CB2  _  1 
CiB  '  AiC  "  BXA  "  C2B    A2C  '  B2A  ~~     * 


Proof.    1.  AC1-AC2  =  BlA-B2A,  Why? 

and  BAX  •  BA2  =  C\B  •  C2B, 

and  CBl-  CB2  =  AXC  -  A2C. 

2.  .'.  by  axs.  6  and  7,  the  above  result  follows. 

Note.  This  theorem,  known  as  Carnot's  theorem,  is  not  a  proposi- 
tion in  concurrence  or  collinearity.  It  is  introduced  as  leading  to  the 
proof  of  the  very  celebrated  theorem  following,  one  commonly  known  as 
the  Mystic  Hexagram,  discovered  by  Pascal  at  the  age  of  16. 

The  theorem  is  also  easily  proved  when  the  triangle  is  inscribed  or 
circumscribed. 


Exercises.     505.    By  means  of  Ceva's  theorem,  prove  that  the  three 
medians  of  a  triangle  are  concurrent. 

506.  Also,  that  the  bisectors  of  the  three  interior  angles  of  a  triangle 
are  concurrent. 

507.  Also,  that  the  bisectors  of  two  exterior  and  of  the  other  interior 
angles  of  a  triangle  are  concurrent. 

508.  Also,  that  the  perpendiculars  from  the  vertices  of  a  triangle  to 
the  opposite  side  are  concurrent. 


242 


PLANE   GEOMETRY. 


[Bk.  V. 


Proposition  XXXV. 

327.    Theorem.     If  the  opposite  sides  of  an  inscribed  hex- 
agon intersect,  they  determine  three  collinear  points. 


Given 


an  inscribed  hexagon,  ABCDEF,  such  that  BA  and 
DE  meet  at  P,  CD  and  AF  at  Q,  BC  and  FE  at  R. 


To  prove    that  P,  Q,  R  are  collinear. 

Proof.    1.  Call  the  A  determined  by  AB,  CD,  and  EF,  LMX. 
as  in  the  figure. 

Then  from  Menelaus's  theorem, 
LP     MD     NE 


and 
and 


PM    DN 

MQ     NF 


QN 
NR 
RL 


FL 
LB 
BM 


EL 

LA 
AM 
MC 

CN 


=  -1, 
=  -1, 
=  -1. 


2.  .'.  By  multiplying  and  recalling  Cam  of s  theorem, 

LP     MQ     NR 
PM  '  QN  '  RL~ 

3.  .'.  by  Menelaus's  theorem,  P,  Q,  R  are  collinear. 


Exs. 509-521.]      CONCURRENCE  AXD   COLLINEARITY.         243 

MISCELLANEOUS  EXERCISES. 

509.  Show  that  the  following  is  a  special  case  of  prop.  XXXI:  The 
perpendicular  bisectors  of  the  sides  of  a  triangle  are  concurrent. 

510.  Also,  the  perpendiculars  from  the  vertices  of  a  triangle  to  the 
opposite  sides  are  concurrent. 

511.  If  three  circumferences  intersect  in  pairs,  the  common  chords 
are  concurrent. 

512.  By  means  of  Menelaus's  theorem,  prove  that  the  points  in  which 
the  three  bisectors  of  the  exterior  angles  of  a  triangle  meet  the  opposite 
sides  are  collinear. 

513.  Also,  that  the  points  in  which  the  two  bisectors  of  two  interior 
angles  of  a  triangle  and  the  other  exterior  angle  meet  the  opposite  sides 
are  collinear. 

514.  The  orthocenter,  0,  of  A  ABC  is  determined  by  the  perpendicu- 
lars AD,  BE.     Prove  that  AO  •  01)  =  BO  ■  OE. 

515.  Draw  a  circle  with  a  central  right  angle  ^4  05.  A  and  B  being 
on  the  circumference;  bisect  ZAOB  by  031,  meeting  AB  at  31;  draw 
MP  __  OA ;  then  see  if  the  following  is  true  in  general :  AB  =  chord 
AB  +  PA.     (Consider  special  cases,  AB  =  120°,  180°,  360°.) 

516.  Given  the  base  and  the  vertical  angle  of  a  triangle ;  construct  it 
so  that  its  area  shall  be  a  maximum. 

517.  AB  is  a  diameter  of  a  circle  of  center  O ;  from  any  point  P  on 
the  circumference,  PC  is  drawn  perpendicular  to  AB)  from  C  a  perpen- 
dicular CE  is  drawn  to  OP.  Prove  that  PC  is  a  mean  proportional 
between  OA  and  PE. 

518.  On  side  a  of  A  ABC,  point  P  is  taken  such  that  Z  PAC  =  Z  B. 
Prove  that  CP  .  CB  =  AP2  :  AB~.  Investigate  for  three  cases,  ZA<, 
=  ,>ZB. 

519.  ABC  is  a  triangle  right-angled  at  C ;  CD±c.  Prove  that 
AD-.LB  =  CA--.BC2. 

520.  If  O,  C  are  the  centers  of  two  fixed  circles,  such  that  the  cir- 
cumference of  C  passes  through  O.  and  if  a  tangent  to  circumference  of 
0  at  T  cuts  circumference  of  0"  at  X,  Y.  then  OX  ■  OY  is  constant.  (If 
the  center-line  meets  the  circumference  of  C at  A,  A  XTO  <-"  A  AYO.) 

521.  If  0  is  the  orthocenter  of  triangle  ABC,  and  A',  B' ',  C  are  the 
mid-points  of  a.b,c;  3Ia.  3Th.  3IC  are  the  mid-points  of  AO,  BO,  CO; 
Pa,  P}j,  Pc  are  the  feet  of  the  perpendiculars  from  A.  B,  C  to  a.  b.  c ; 
prove  that  A\  B' .  C",  Ma,  3Ib.  3E.  Pa,  Pb,  Pc  are  coney clic.  (The 
••  Nine  Points  Circle.") 


SOLID   GEOMETRY. 


BOOK   VI.  — LINES   AND   PLANES   IN   SPACE. 


1.      THE  POSITION  OF  A   PLANE  IN  SPACE.     THE  STRAIGHT 
LINE   AS    THE   INTERSECTION   OF   TWO   PLANES. 

328.    Definitions.     Through  three  points,  not  in  a  straight 
line,  any  number  of  surfaces  may 
be  imagined  to  pass. 

For  example,  through  the  points  A,  B,     _/_ A^ 

C  the  surfaces  P  and  S  may  be  imagined 
to 


329.  A  plane  surface  (also  called  &  plane)  is  a  surface  which 
is  determined  by  any  three  of  its  points  not  in  a  straight  line. 

In  the  figure,  P  represents  a  plane,  for  it  is  determined  by  the  points 
A,  B,  C.     But  S  does  not  represent  such  a  surface. 

A  plane  is,  of  course,  supposed  to  be  indefinite  in  extent. 

This  definition,  and  the  following  postulates,  are  repeated,  for  con- 
venience, from  the  Plane  Geometry. 

In  drawing  a  figure  it  should  be  remembered  that  a  plane, 
like  a  line,  has  no  thickness,  and  that  it  is  indefinite  in  extent. 
Nevertheless,  it  aids  the  eye  in  understanding  the  figure,  if  we 
represent  the  plane  as  a  rectangle,  lying  in  perspective,  and 
having  a  slight  thickness. 


Exercises.  522.  Show  that  if  there  are  given  four  points  in  space, 
no  three  being  collinear,  the  number  of  distinct  straight  lines  determined 
by  them  is  six  ;  if  there  are  five  points,  the  number  of  lines  is  ten. 

523.  Hold  two  pencils  in  such  a  way  as  to  show  that  a  plane  cannot, 
in  general,  contain  two  straight  lines  taken  at  random  in  space. 

244 


Prop.  I.]  LINES   AND   PLANES  IN  SPACE.  245 

330.  Postulates  of  the  Plane.     (See  §  29.) 

1.  Three  points  not  in  a  straight  line  determine  a  plane. 

2.  A  straight  line  through  two  points  in  a  plane  lies  wholly 
in  the  plane. 

3.  A  plane  may  be  pjassed   through    a    straight   line  and 
revolved  about  it  so  as  to  include  any  assigned  point  in  space. 

4.  A  portion  of  a  plane  may  be  produced. 

5.  A  plane  is   divided   into   two  parts   by  any  one  of  its 
straight  lines,  and  space  is  divided  into  two  parts  by  any  plane. 

331.  Solid  Geometry  treats  of  figures  whose  parts  are  not 
all  in  one  plane. 

Proposition  I. 

332.  Theorem.     A  plane  is  determined  by  a  straight  line 
and  a  point  not  in  that  line. 


Given         the  line  AB,  and  the  point  P  not  in  that  line. 

To  prove    that  AB  and  P  determine  a  plane. 

Proof.    1.  Only  one  plane  contains  pts.  A,  B,  and  P.    §  330,  1 
(§  330,  1.     Three  points  not  in  a  straight  line  determine  a  plane.) 

2.  And  that  plane  contains  line  AB.  §  330,  2 

(§  330,  2.     A  straight  line  joining  two  points  in  a  plane  lies  wholly 
in  the  plane.) 

3.  .*.  only  one  plane  contains  AB  and  P. 

333.    Definition.     Lines  or  points  which  lie  in  the  same  plane 
are  said  to  be  coplanar. 


246 


SOLID   GEOMETRY. 


[Bk.  VI. 


Corollaries.     1.  A  plane  is  determined  by  two  intersecting 
lines. 


Let  the  lines  AB,  CD  intersect  at  0. 

Then  only  one  plane  contains  AB  and  C.  Prop.  I 

And  that  plane  contains  the  point  0,  for  0  lies  in  the  line  AB. 

§  330,  2 
And  since  that  plane  contains  C  and  O,  it  contains  CD.  §  330,  2 

2.  A  plane  is  determined  by  tiuo  parallel  lines. 

For  the  parallels  lie  in  one  plane,  by  definition  (§  82). 
And  only  one  plane  can  contain  these  parallels,  since  a  plane  is  deter- 
mined by  either  line  and  any  point  of  the  other. 
Draw  the  figure. 

3.  If  a  plane  contains  one  of  two  ])arallel  lines  and  any 
point  of  the  other,  it  contains  both  parallel  lines. 

For  it  must  be  identical  with  the  plane  determined  by  the  two  paral- 
lels ;  otherwise  more  than  one  plane  could  contain  either  parallel  and  any 
point  in  the  other. 


Proposition  II. 

334.    Theorem.     The  intersection  of  two  planes  is  a  straight 

line. 

/N 


Given         two  intersecting  planes,  M,  N. 

To  prove    that  their  intersection  is  a  straight  line. 


Prop.  II.]  LINES  AND  PLANES  IN  SPACE.  247 

Proof.    1.  Let  P  be  a  point  common  to  M  and  N. 

Then  a  pencil  of  lines  through  P,  in  the  plane  X, 
must  lie  partly  on  one  side  of  21  and  partly  on  the 
other,  because  21  divides  space  into  two  parts. 

§  330,  5 

2.  Hence,  in  general,  a  line  connecting  a  point  in  the 
pencil  on  one  side  of  21,  with  a  point  on  the  other 
side,  must  cut  21  at  some  other  point  than  P,  — 
say  at   Q. 

3.  Then  21  and  X  have  two  points  in  common. 

4.  Then  every  point  in  the  straight  line  through  P  and 
Q  lies  in  plane  21,  §  330,  2 
and  also  in  plane  X,  for  the  same  reason. 

5.  .'.  the  straight  line  PQ  is  common  to  both  planes. 

6.  If  there  were  any  point  not  in  PQ,  common  to  21 
and  X,  the  planes  would  coincide.  Prop.  I 

Corollary.     A  point  common  to  two  planes  lies  in  their 
line  of  intersection. 
Proved  in  step  6. 

Exercises.     524.    State    the    four    methods,    already   mentioned,    of 
determining  a  plane. 

525.  Is  it  possible  for  three  planes  to  have  a  straight  line  in  common  ? 
Draw  a  figure  to  illustrate. 

526.  If  two  planes  have  three  points  in  common,  will  they  necessarily 
coincide  ? 

527.  Four  planes,  no  three  containing  the  same  line,  intersect  in  pairs  ; 
how  many  straight  lines  do  they  determine  by  their  intersections  ? 

528.  What  is  the  only  rectilinear  polygon  that  is  necessarily  plane  ? 
Why? 

529.  Prove  that  all  transversals  of  two  parallel  lines  are  coplanar  with 
the  parallels. 

530.  What  is  the  reason  that  a  three-legged  chair  is  always  stable  on 
the  floor  while  a  four-legged  one  may  not  be  ? 


248  SOLID   GEOMETRY.  [Bk.  VI. 

Proposition  III. 

335.  Theorem.  If  three  planes,  not  containing  the  same 
line,  intersect  in  pairs,  the  three  lines  of  intersection  are 
either  concurrent  or  parallel. 


Given         planes  AD,  CF,  EB,  intersecting  in  AB,  CD,  EF. 

To  prove    that  AB,  CD,  EF  are  either  concurrent  or  parallel. 

Proof.    Case  I.     If  CD  meets  AB,  as  at  0,  to  show  the  three 
lines  concurrent. 

1.  v  0  is  in  AB,  it  is  in  plane  EB.  §  330,  2 

2.  Similarly,  V  0  is  in  CD,  it  is  in  plane  CF.      Why  ? 

3.  V  0  is  in  planes  EB  and  CF,  EF  passes  through  0. 

Prop.  II,  cor. 

4.  .'.  AB,  CD,  EF  are  concurrent  in  0. 

Case  II.     If  CD  II  AB,  to  show  the  three  lines  parallel. 

1.  If  AB  were  not  II  EF,  CD  would  pass  through  their 
common  point.  Case  I 

2.  But  this  is  impossible,  for  CD  II  AB.  Given 

3.  If  CD  were  not  II  EF,  AB  would  pass  through  their 
common  point.  Case  I 

4.  But  this  is  impossible,  for  CD  II  AB.  .    Given 

5.  .*.  as  no  two  can  meet,  and  as  each  pair  is  coplanar, 
they  are  parallel.  Def.  II  lines 

Corollary.     If  tivo  intersectl»<j  planes  pass  through  two 

parallel  lines,  their  intersectio)/   is  parallel  to  these  lines. 


Prop.  IV.]  LINES   AND   PLANES  IN  SPACE.  249 

Proposition  IV. 

336.    Theorem.     Lines  parallel  to  the  same  line  are  parallel 
to  each  other. 


Given  AB  II  EF,  CD  II  EF. 

To  prove    that  AB  II  CD. 

Proof.    1.  AB  and  EF  determine  a  plane.  Prop.  I,  cor.  2 

(A  plane  is  determined  by  two  II  lines.) 

2.  CD  and  EF  determine  a  plane.  Why  ? 

3.  AB  and  any  point  C  of  CD  determine  a  plane. 

Prop.  I 

4.  Suppose  this  last  plane  to  intersect  plane  ED  in  CX, 
another  line  than  CD. 

Then  CX  would  be  II  to  both  EF  and  AB. 

Prop.  Ill,  cor. 
(If  two  intersecting  planes  pass  through  two  II  lines,  their  intersection 
is  II  to  these  lines.) 

5.  But  v  CD  II  EF,  this  is  impossible. 

Post,  of  parallels 

(§  85.    Two  intersecting  straight  lines  cannot  both  be  II  to  the  same 
straight  line.) 

6.  .'.  CD  is  the  intersection  of  the  planes  through  AB 
and  C,  and  EF  and  C, 

and  .'.CD  II  AB.  Prop.  Ill,  cor. 


Exercise.     531.    Why  will  not  the  proof  of  this  theorem  as  given  in 
plane  geometry  apply  to  this  case  in  solid  geometry  ? 


250  SOLID   GEOMETRY.  [Bk.  VI. 

Proposition   V. 

337.  Theorem.  If  two  intersecting  lines  are  respectively 
parallel  to  two  others,  the  angles  made  by  the  first  pair  are 
equal  or  supplemental  to  those  made  by  the  second  pair. 


Given  two  intersecting  lines  x,  y,  respectively  parallel  to 
two  other  lines  x',  if. 

To  prove  that  the  angles  made  by  x  and  y  are  equal  or  supple- 
mental to  those  made  by  x'  and  f. 

Proof.  1.  Suppose  the  intersections  0  and  0'  are  joined,  and 
from  any  points  A,  B,  on  x,  y,  parallels  to  00'  are 
drawn. 

2.  '.'00',  x,  and  x'  are  coplanar  (Why?),  the  parallel 
from  A  meets  x'  as  at  A'.     Similarly,  B'  is  fixed. 

Prop.  I,  cor.  3 

3.  Draw  AB,  A'B'. 

'.'  AA'  II  00',  and  BB'  II  00',  .'.  AA'  II  BB'.     Prop.  IV 

4.  '.•OA,,OB'axeU],.\AA'=00'  =  BB'.  I,  prop.  XXIV 

5.  .-.  ABBA'  is  a  O.  I,  prop.  XXV 

6.  .'.  OA  =  O'A',  OB  =  O'B',  AB  =  A'B'.  I,  prop.  XXIV 

7.  .'.  A  ABO  S  A  A'B'O',  and  Z  AOB  =  Z  A' O'B'. 

I,  prop.  XII 
§  After  proving  one  pair  of  angles  equal,  the  rest  are 
evidently  equal  or  supplemental  by  the  theorems  con- 
cerning vertical  and  supplemental  angles. 


Prop.  VI. 


LINES  A  XI)   PLANES   IX  SPACE. 


251 


2.     THE   RELATIVE   POSITION   OF   A   LINE   AND   A   PLANE. 
Proposition  VI. 

338.  Theorem.  If  a  line  is  perpendicular  to  each  of  two 
intersecting  lines,  it  is  perpendicular  to  every  other  line 
lying  in  their  plane  and  passing  through  their  point  oj 
intersection. 


Given  x  and  z,  two  lines  intersecting  at  0,  and  w  perpen- 
dicular to  x  and  to  z ;  also  y,  any  line  through  0 
coplanar  with  x,  z. 

To  prove    that  w  _L  y. 

Proof.  1.  On  w  suppose  OP'  =  PO  ;  let  any  transversal  cut 
x,  y,  z  at  A,  B,  C ;  join  P  and  P'  with  A,  B,  C. 

Then  AP  =  AP',  and  CP  =  CP'.    I,  prop.  XX,  cor.  5 

2.  And  '.-  AC  =  AC, 

.-.AACP^A  A  CP'.  I,  prop.  XII 

3.  .'.by  folding  A  ACP  over  AC  as  an  axis,  it  can  be 
brought  to  coincide  with  A  A  CP'.  §  57 


4.  /.A  BOP  ^  A  BOP', 

and  Z  POB  is  a  rt.  Z,  and  w  J_  y. 


I,  prop.  XII 
Why  ? 


9m 


SOLID   GEOMETRY. 


[Bk.  VI. 


339.  Definitions.  A  line  is  said  to  be  perpendicular  to  a  plane 
when  it  is  perpendicular  to  every  line  in  that  plane  which 
passes  through  its  foot,  —  i.e.  the  point  where  it  meets  the 
plane.     The  plane  is  then  said  to  be  perpendicular  to  the  line. 

If  a  line  meets  a  plane,  and  is  not  perpendicular  to  it,  it  is 
said  to  be  oblique  to  the  plane. 

Corollaries.  1.  If  a  line  is  perpendicular  to  each  of  two 
intersecting  lines,  it  is  perpendicular  to  their  plane. 

2.  The  locus  of  points  equidistant  from  two  given  joints  is 
the  plane  bisecting  at  right  angles  the  line  joining  those  points. 


Proposition  VII. 

340.  Theorem.  If  a  line  is  perpendicular  to  each  of  three 
concurrent  lines  at  their  point  of  concurrence,  the  three  lines 
are  coplanar. 


Given  OY±OA,  OB,  OC. 

To  prove    that         OA,  OB,  OC  are  coplanar. 

Proof.    1.   Suppose  M  the  plane  determined  by  OA,  OB;   and 
N  the  plane  determined  by  OY,  OC. 

Suppose  that  OC  is  not  in  M,  and  call  OX  the  inter- 
section of  M  and  N. 


Prop.  VII.]         LINES   AX  I)    PLANES    IN   SPACE.  253 

2.  Then  must  OY ±  OX.  Prop.  VI 

3.  But  v  OY  ±  OC,  this  is  impossible.      Prel.  prop.  II 

4.  .'.  it  is  absurd  to  suppose  OC  not  in  M  with  OA 
and  OB. 

Corollaries.    1.  Lines  perpendicular  to  the  same  line  at 
the  same  point  are  coplanar. 

2.    Through  a  given  point  in  a  -plane  there  cannot  be  drawn 
more  than  one  line  perpendicular  to  that  plane. 


Suppose  OP  and  OQ  _L  plane  M.  Then  each  would  be  perpendicular 
to  QX,  the  line  of  intersection  of  their  plane  N  with  the  given  plane  M, 
thus  violating  prel.  prop.  II. 

3.  Through  a  given  point  in  a  line  there  cannot  be  drawn 
more  than  one  plane  perpendicular  to  that  line. 

For  if  two  planes  could  be  drawn  perpendicular  to  the  line,  then  three 
lines  in  each  would  be  perpendicular  to  the  given  line,  and  hence  the  two 
planes  would  coincide. 

Exercises.  532.  Prove  that  if  the  hand  of  a  clock  is  perpendicular 
to  its  moving  axle,  it  describes  a  plane  in  its  revolution.  Prove  the 
converse. 

533.  How  many  straight  lines  are  determined  by  six  points,  three 
being  collinear  ? 

534.  How  many  planes  in  general  are  determined  by  four  points  in 
space,  no  three  being  collinear  ? 

535.  In  the  left-hand  figure  of  prop.  Ill,  suppose  point  0  to  move 
farther  from  BDF,  and  to  continue  to  do  so  indefinitely.  What  is  the 
limiting  figure  which  the  left-hand  figure  is  approaching? 


254 


SOLID   GEOMETRY. 


[Bk.  VI. 


Proposition   VIII. 

341.    Theorem.     Lines  perpendicular  to  the  same  plane  are 
parallel. 

A 


Given  0  Y,  XA  _L  plane  MN  at  0,  X. 

To  prove    that  0  Y  II  XA. 

Proof.         It   is   necessary  first  to    show   that    0  Y,  XA  are 
coplanar ;  then  that  they  are  _L  to  OX. 

1.  Let  XZ±  OX  in  plane  MN,  and  =  OY. 
Draw  OZ,  ZY,  XY. 
v  XZ=  OY,  OX=  OX, 
ZXOY=ZOXZ=vt.Z, 
.AXOY^A  OXZ, 

OZ  =  XY.  I,  prop.  I 

\'ZY=ZY,  .'.AXYZ^AOZY, 
Z  YXZ  =  Z.ZOY=  rt.  Z.  I,  prop.  XII 

'.  XA,  XY,  XO  are  coplanar.  Why  ? 

'.  YO  lies  in  that  same  plane.  §  330,  2 

v  YO  and  AX  ±  OX,  §  339 

.*.  YO  II  AX,  and  similarly  for  all  other  Js. 

I,  prop.  XVI,  cor.  3 

Corollary.     Front  a  point  outside  of  a  ])la?ie,  not  more 
than  one  line  can  be  drawn  perpendicular  to  that  plane. 


2. 

Then 

o 

and 

.). 

and 

4. 

And 

and 

5. 

6. 

But 

Prop.  IX.] 


LINES  AND  PLANES   IN  SPACE. 


255 


Proposition  IX. 

342.   Theorem.     If  one  of  two  parallel  lines  is  perpendicu- 
lar to  a  plane,  the  other  is  also. 


Given 


OY  II  O'Y',  OF _L  plane  Jf2Vat  0,  and  O'Y'  meeting 
plane  3IJY  at  0'. 


To  prove    that 


0'Y'±MN. 


Proof.    1.  Let  OA,  OB  be  any  lines  from  0,  in  MN,  O'A'  II  OA, 
and  O'B'  II  OB. 

2.  Then        A  YOA,  YOB  are  rt.  A.  §  339 

3.  But  Z  YOA  =  Z  Y'O'A',  Z  YOB  =  Z  Y'O'B'.    Prop.  V 

4.  .'.  A  Y'O'A',  Y'O'B'  are  also  rt.  A,  Prel.  prop.  I 
and                     O'Y'A-MN.  §339 

343.  Definitions.  The  projection  of  a  point  on  a  plane  is  the 
foot  of  the  perpendicular  through  that  point  to  the  plane. 

The  projection  of  a  line  on  a  plane  is  the  locus  of  the  pro- 
jections of  all  of  its  points. 


Exercises.  536.  Are  lines  which  make  equal  angles  with  a  given 
line  always  parallel  ?     (Answer  by  drawing  figures  to  illustrate.) 

537.  Show  how  to  determine  the  perpendicular  to  a  plane,  through  a 
given  point,  by  the  use  of  two  carpenter's  squares. 

538.  Prove  prop.  VI  on  the  following  outline  :  Assume  B  on  y,  and 
draw  ABC  so  that  AB  =  BC  (How  is  this  done?);  prove  2PB24- 
2  •  BC2  =  PA2  +  PC2  =  2  •  PO2  +  OC2  +  OA2  =  2  -PO2  +  2  ■  OB2  + 
2  •  BC2 ;  .-.  PB2  =  PO2  +  OB2;  .-.  Z  POB  is  a  rt.  Z. 


256 


SOLID    GEOMETRY. 


[Bk.  V 


Proposition  X. 

344.  Theorem.  The  projection  of  a  straight  line  on  a  plane 
is  the  straight  line  which  jiasses  through  the  projections  of 
any  two  of  its  points. 


Given        A',  P',  B',  the  projections  of  A,  P,  B,  points  in  the 

line  AB,  on  the  plane  MN. 
To  prove   that  the  projection  of  AB  is  the  straight  line  A'B'. 


Why  ? 
Prop.  I,  cor.  2 

Why  ? 

Prop.  I,  cor.  3 

Prop.  II 


Proof.  1.  AA'  II  BB'  II  PP'. 

2.  .'*.  A,  A',  B,  B'  are  coplanar. 

3.  .'.  P  is  in  that  same  plane. 

4.  .*.  PP'  is  in  that  same  plane. 

5.  .'.A',  P',  B'  are  collinear. 

6.  Also  any  other  point  in  A'B'  is  the  projection  of  some 
point  in  AB.  For  a  _L  to  MN  from  such  a  point 
is  II  to  AA'  (§  341)  and  lies  in  plane  AA'B'B  (§  82), 
and  therefore  meets  AB  in  some  point. 

345.  Definitions.  The  smallest  angle  formed  by  a  line  and 
its  projection  on  a  plane  is  called  the  inclination  of  the  line  to 
the  plane  or  the  angle  of  the  line  and  the  plane. 

A  figure  is  said  to  be  projected  on  a  plane  when  all  of  its 
points  are  projected  on  the  plane. 

The  plane  determined  by  a  line  and  its  projection  on  anot  her 
plane  is  called  the  projecting  plane. 


In  the  figure  of  pro]).  X,  Z  B'OB  is  the  incjination  of  A  B  to  MN.    The 
plane  determined  by  AB,  A'T¥  is  the  projecting  plane. 


Prop.  XL]  LINES   AND   PL  AXES   IN  SPACE.  257 

Proposition  XL 

346.    Theorem.      Of  all  lines   that  can  be   drawn  from  a 
point  to  a  plane, 

1.  The  perpendicular  is  the  shortest; 

2.  Obliques  with  equal  inclinations  are  equal,  and  con- 
versely ; 

3.  Obliques  with   equal  projections   are   equal,  and   con- 
versely. 


1.  Given        PO  J_  plane  MN,  PX  oblique  to  MN. 
To  prove  that  PO  <  PX. 

Proof.  V  Z  XOP  =  rt.  Z.  (Why?)  .\PO<  PX.    I,  prop.  XX 

2.  Given  PO  ZJIX.  ZPYO  =  ZPXO. 

To  prove    that  PY=PX,  which   is   true  because  APOY  = 
A  POX.  I,  prop.  XIX,  cor.  7 

Conversely  :  Given    PO  _L  MN,  PY=PX. 

To  prove    that    ZPYO  =  ZPXO,    which    is    true    because 
A  POT  ^  A  POX.  I,  prop.  XIX,  cor.  5 

3.  Given  PO  ±  JIX,  OY=  OX. 

To  prove    that  PY=PX,   which   is   true  because  APOY^ 
A  POX.  I,  prop.  I 

Conversely  :  Given  PO  _1_  JIX,  PY=PX. 

To  prove    that    OY=OX,   which   is   true   because  APOY^ 
A  POX.  I,  prop.  XIX,  cor.  5 


258  SOLID   GEOMETRY.  [Bk.  VI. 

Proposition  XII. 

347.    Theorem.     From  a  point  to  a  plane, 

1.  Of  two  obliques  with  unequal  inclinations,  that  having 
the  greater  inclination  is  the  shorter,  and  conversely  ; 

2.  Of  two  obliques  with  unequal  projections,  that  having 
the  longer  projection  is  the  longer,  and  conversely. 


1.  Given    PO  _L  MN9  PY  and  PA  two  obliques   such  that 
Z.PAO>  APYO. 

To  prove    that  PA<PY. 

Proof.    1.  Suppose  X  taken  on  OA  so  that  OX  =  0  Y. 

2.  Then  APOX^APOY,  and  PX  =  P  Y.        Why  ? 

3.  But  PX,  and  .-.its  equal  PY,  >  PA.      I,  prop.  XX 
Conversely  : 

Given         PO  ±  MX,  PY  and  PA   two  obliques  such  that 
PA  <  PY. 

To  prove    that  Z  PAD  >  Z  P  YO. 

Proof.    1.  Suppose  X  taken  on  OA  so  that  OX  =  0  Y. 

2.  Then   A  POX  ^  A  PO  Y,    Z  PXO  =ZPYO,    and 
PX=PY.  Why? 

3.  ..PA<  PX,  '.'  PA  <PY.  Given 

4.  X  cannot  fall  on  A,  for  then       PA  =  PA'. 

5.  Nor  between  0  and  A,  for  then  PA  >  PX.     Why? 


Prop.  XII.]         LINES   AND   PLANES   IN  SPACE.  259 

6.  .'.  X  is  on  OA  produced, 

and  .'.ZPAO>  Z  PXO.  I,  prop.  V 

7.  .'.  Z  PJ 0  >  Z  P  FO.  Subst. 

2.  Given    PO  J_  MN,  PY  and  P.4  two  obliques   such  that 
OA  <  OY. 

To  prove    that  PA<PY. 

Proof.    1.   Suppose  X taken  on  OA  so  that  OX  =  OY. 

2.  Then  APOX^APOY,  and  PX=  PY.     I,  prop.  I 

3.  And  v  (XI  <  OF,  or  OX, 

.*.  P.4  <  PX,  or  ?7.  Why  ? 

Conversely : 
Given         PO  _L  MN,  PY  and  PA  two  obliques  such  that 
PA  <  PY. 

To  prove    that  OA  <  OY. 
Proof  left  for  the  student. 

Definition.  The  length  of  the  perpendicular  from  a  point  to 
a  plane  is  called  the  distance  from  that  point  to  the  plane. 

E.g.  in  the  figure  on  p.  258,  the  distance  from  P  to  MN  is  the  length 
of  PO.  

Exercises.  539.  Prove  that  if  three  concurrent  lines  meet  a  fourth 
line,  not  in  the  same  point,  the  four  lines  are  coplanar. 

540.  Why  does  folding  a  sheet  of  paper  give  a  straight  edge  ? 

541.  Suppose  it  known  that  a  point  P  is  in  each  of  the  three  planes 
X,  Y,  Z.     Is  P  probably  fixed  ?     Is  it  necessarily  fixed  ? 

542.  If  the  triangles  ABC,  A'B'C,  in  different  planes,  are  such  that 
AB  and  A'B'  meet  when  produced,  as  also  BC  and  B'C,  and  CA  and 
C'A\  then  the  lines  AA\  BB\  CC  are  either  concurrent  or  parallel. 

543.  How  many  planes  are  determined  by  n  concurrent  lines,  no  three 
of  which  are  coplanar  ? 

544.  If  a  line  cuts  one  of  two  parallel  lines,  must  it  cut  the  other  ?  If 
it  does,  are  the  corresponding  angles  equal  ? 


260  SOLID   GEOMETRY.  [Bic.  VI. 

Proposition  XIII. 

348.  Theorem.  The  acute  angle  which  a  line  makes  with 
its  own  projection  on  a  plane  is  the  least  angle  ivhich  it  makes 
with  any  line  in  that  plane. 


ft. 

1 

%\ 

1 

1  B' 

0^- 

~^_^    J 

a 

?L 

1r 

\ 

Given  the  line  AB,  cutting  plane  P  at  0,  A'B'  the  projec- 
tion of  AB  on  P,  and  XX'  any  other  line  in  P, 
through  0. 


'oJ 


To  prove    that  Z  A'OA  <  Z  XOA. 

Proof.    1.   Suppose  A'  the  projection  of  A,  OX  made  equal  to 
OA',  and  AX,  AA'  drawn. 

2.  Then  AA'  <  AX.  Prop.  XI,  1 

3.  .'.  in  A  OX  A  and  OA  'A,  we  have 

OX=  OA',   OA  =  OA,  and  AA'  <  AX, 

.-.  Z  A'OA  <  Z  XOA.  I,  prop.  XI 

Exercises.  545.  The  obtuse  angle  which  a  line  makes  with  its  own 
projection  (produced)  on  a  plane  is  the  greatest  angle  which  it  makes 
with  any  line  in  that  plane. 

546.  In  a  given  plane,  to  determine  the  locus  of  points  equidistant 
from  two  fixed  points  in  space. 

547.  Prove  prop.  IX  by  supposing  O'Y'  not  perpendicular  to  MN,  but 
supposing  another  line  O'Z  from  0'  _L  to  MN  ;  then  prove  that  O'Z  would 
be  parallel  to  OF,  which  would  violate  §  85,  and  hence  be  absurd. 

548.  As  a  special  case  of  prop.  X.  suppose  AB  ±  MN ;  what  would 
be  its  projection  ami  its  inclination  ? 


Prop.  XIV.]        LINES   A  XL)   PL  AXES    IX   SPACE.  2G1 

Proposition  XIV. 

349.    Theorem.     Parallel  lines  intersecting  the  same  plane 
are  equally  inclined  to  it. 

?' 


M 
Given         two  parallels.  PA,  P'A',  intersecting  a  plane  MN  at 
A,  A' ;   and  0,  0'  the  projections  of  P,  P. 

To  prove    that  Z  PAD  =  Z  P'A'O'. 

Proof.    1.  vP0andP'0'_LJAV. 

.-.POWP'O'.  Why? 

2.  .-.  Z  OP  A  =  Z  OP' A'.  Prop.  V 

(Let  the  student  complete  the  proof.) 

350.  Definition.  Two  straight  lines,  not  coplanar,  are  re- 
garded as  forming  an  angle  which  is  equal  to  the  one  formed 
by  either  line  and  a  line  drawn,  from  a  point  upon  it,  parallel 
to  the  second. 

E.g.  in  the  figure  of  prop.  XIV.  the  angle  made  by  AO  and  P'A'  is 
considered  as  Z  P'A' (J  or  Z  PAD. 


t 


Exercises.  549.  Parallel  line-segments  are  proportional  to  their  pro- 
j < - e t i <  »ns  on  a  plane. 

550.  In  general,  which  is  the  longer,  a  line-segment  or  its  projection  ? 
Is  there  any  exception  ? 

551.  Show  how,  with  a  10  ft.  pole  marked  in  feet,  to  determine  the 
foot  of  the  perpendicular  let  fall  to  the  floor  from  the  ceiling  of  a  room 
8  ft.  high. 

552.  Show  how  a  line  1  in.  long  and  another  2  in.  long  may  have 
equal  projections  on  a  plane. 

553.  If  any  two  lines  are  parallel,  respectively,  to  two  others,  an  angle 
made  by  the  first  pair  equals  one  made  by  the  second. 


262 


SOLID   GEOMETRY. 


[Bk.  VI. 


Proposition  XV. 

351.  Theorem.  If  a  line  intersects  a  plane,  the  line  in  the 
plane  perpendicular  to  the  projection  of  the  first  line  at  the 
point  of  intersection  is  perpendicular  to  the  line  itself 


Given         AB  intersecting  the  plane  MN  at  A,  B'  the  projec- 
tion of  B  on  MN,  and  DC  _L  AB1  at  A. 

To  prove    that  DC  A.  AB. 

Proof.    1.  On  DC  let  EA  =  AE' ;   join  E,  E'  to  B  and  B'. 

2.  Then  A  AB'E  ^  A  AB'E',  and  EB'  =  E'B'.     Why  ? 

3.  Then  A  EBB'  ^  A  E'BB',  and  EB  =  E'B.       Why  ? 

4.  Then  A  ^'^Z?  ^  A  JE45,  and  Z  E'AB  =  Z  £^j£ 

Why  ? 

5.  .\DC  ±AB,  by  defs.  of  rt.  Z  and  J_. 

352.  Definition.  A  line  is  said  to  be  parallel  to  a  plane  when 
it  never  meets  the  plane,  however  far  produced.  In  that  case, 
also,  the  plane  is  said  to  be  parallel  to  the  line. 


Exercises.  554.  Prove  prop.  XV  on  the  following  outline :  draw 
through  B'  a  line  II  to  DC ;  prove  this  parallel  perpendicular  to  plane 
AB'B  ;  .-.  DC  ±  plane  AB'B,  .:  DC  ±  AB. 

555.  Prove  prop.  XV  by  showing  that  AE"2  -f  AB2  =  BE'2,  .and  that 
therefore  Z  E'AB  is  right. 

556.  In  the  figure  of  prop.  XV  prove  that  the  area  and  perimeter  of 
I\AB'B  are  respectively  less  than  those  of  AEB'B. 


Prop.  XVI.]        LINES   AND   PLANES   IN  SPACE.  263 

Proposition  XVI. 

353.    Theorem.     Any  plane  containing  only  one  of  two  par- 
allel lines  is  parallel  to  the  other. 


Given         the  parallel  lines  AB,  A'B',  and  the  plane  MN  con- 
taining AB  but  not  A'B'. 

To  prove    that  MN  II  A'B'. 

Proof.    1.  AB  and  A'B'  determine  a  plane  P.      Prop.  I,  cor.  2 

2.  v  AB  and  A'B'  lie  wholly  in  P,  .-.  if  ^'^'  meets  JlfiV 
it  meets  AB.  §  85 

3.  But  v  AB  II  A'B',  this  is  impossible.  §  82,  del  II  lines 


Exercises.  557.  A  line  which  is  parallel  to  a  plane  is  parallel  to  its 
projection  on  that  plane. 

558.  Through  a  point  without  a  straight  line  any  number  of  planes 
can  pass  parallel  to  that  line. 

559.  If  a  line  is  parallel  to  a  plane,  the  intersection  of  that  plane  with 
any  plane  passing  through  that  line  is  parallel  to  the  line. 

560.  If  from  two  points  on  a  line  parallel  to  a  plane,  parallel  lines  are 
drawn  to  and  terminated  by  that  plane,  these  parallel  lines  are  equal. 

561.  If  a  line  is  parallel  to  a  plane,  and  if  from  any  point  in  the  plane 
a  line  is  drawn  parallel  to  the  first  line,  then  the  second  line  lies  wholly 
in  the  plane. 

562.  If,  through  a  line  parallel  to  a  plane,  several  planes  pass  so  as  to 
intersect  that  plane,  these  lines  of  intersection  are  parallel. 

563.  If  the  distances  from  two  given  points  on  the  same  side  of  a 
plane,  to  that  plane,  are  equal,  the  line  determined  by  those  points  is 
parallel  to  the  plane. 


264  SOLID   GEOMETRY.  [Bk.  VI. 

Proposition  XVII. 

354.    Theorem.     Between  two  lines  not  in  the  same  plane, 
one,  and  only  one,  common  perpendicular  can  be  drawn. 


M 

Given         two  lines  K,  L,  not  coplanar. 

To  prove    that  one,  and  only  one,  common  perpendicular  can 
be  drawn  between  them. 

Proof.    1.  Let  MN  be  the  plane,  through  K,  II  L.     (Can  such  a 
plane  exist  ?)     Let  L'  be  the  projection  of  L  on  MN. 

2.  Then  K  is  not  II  to  L',  for  then  it  would  be  II  to  L. 

Why  ? 
Let  K  intersect  V  at  P. 

3.  A  _L  to  L  and  K  is  _L  to  MN.  Why  ? 

4.  Then  v  L'  is  the  locus  of  the  feet  of  all  Js  from 
points  in  L,  on  plane  MN,  §  343,  def.  projection 
.".  P  is  the  unique  point  in  which  a  _L  from  a  point 
on  L,  to  K,  can  meet  K. 

5.  .'.  if  PQ  is  drawn  _L  to  L,  it  is  _L,  and  the  only  _L, 
to  both  L  and  A". 

Corollary.     The  common  perpendicular  is  the  shortest  line- 
segment  between  two  lines  not  in  the  same  i>lnne. 

For  if  QfP  II  QP,  then  QP  =  Q'P'  <  QfR.  Prop.  XI,  1 

355.    Definition.     The  length  of  the  common  perpendicular 

from  one  line  to  another  is  called  the  distance  between  those 
lines. 


K 


Sbcs.  350-358.] 


PENCIL    OF  PLANES. 


265 


3.     PENCIL   OF   PLANES. 

356.  Definitions.  Any  number  of  planes  containing  the 
same  line  are  said  to  form  a  pencil  of  planes ;  the  line  is 
called  its  axis. 


357.    Any  two  planes  of  a  pencil  are  said  to  form  a  dihedral 
angle. 


LMN,  a  pencil  of  planes  ;  A B,  the 
axis  of  the  pencil. 


Dihedral  angles  formed  by  the 
planes  M  and  ^V.  Dihedral  angle 
MN  measured  by  plane  angle 
BOC.  AOthe  edge  of  the  dihe- 
dral angle. 


The  two  planes  of  a  dihedral  angle  are  called  the  faces,  and 
the  axis  of  the  pencil  is  called  the  edge  of  the  dihedral  angle. 

Two  intersecting  planes  form  more  than  one  dihedral  angle,  just  as 
two  intersecting  lines  form  more  than  one  plane  angle,  the  latter  term 
now  being  used  to  designate  an  angle  made  by  lines  in  a  plane. 

358.  A  plane  of  a  pencil  turning  about  the  axis  from  one 
face  of  a  dihedral  angle  to  the  other  is  said  to  turn  through 
the  angle,  the  angle  being  greater  as  the  amount  of  turning  is 
greater. 

Since  the  size  of  a  dihedral  angle  depends  only  upon  the 
amount  of  turning  just  mentioned,  it  is  independent  of  the 
extent  of  the  faces. 


266  SOLID   GEOMETRY.  [Bk.  VI. 

359.  If  perpendiculars  are  erected  from  any  point  in  the 
edge  of  a  dihedral  angle,  one  in  each  face,  the  size  of  the  plane 
angle  thus  formed  evidently  varies  as  the  size  of  the  dihedral 
angle.  Hence  a  dihedral  angle  is  said  to  be  measured  by  that 
plane  angle,  or,  strictly,  to  have  the  same  numerical  measure. 

360.  A  dihedral  angle  is  said  to  be  acute,  right,  obtuse, 
oblique,  reflex,  straight,  according  as  the  measuring  plane 
angle  is  so,  and  it  is  usually  named  by  its  measuring  plane 
angle,  or  merely  by  its  faces  in  counter-clockwise  order. 

The  terms  adjacent  angles,  bisector,  sum  and  difference  of 
dihedral  angles,  point  within  or  without  the  angle,  complement, 
supplement,  conjugate,  and  vertical  angles  will  readily  be 
understood  from  the  corresponding  terms  in  plane  geometry. 

As  with  plane  angles  the  smallest  angle  made  by  two  intersecting 
lines  is,  in  general,  to  be  understood  unless  the  contrary  is  stated,  so 
with  dihedral  angles. 

If  a  dihedral  angle  is  right,  the  planes  are  said  to  be  perpen- 
dicular to  each  other. 

E.g.  in  the  following  figure,  ZY  ±  MN.. 


Proposition  XVIII. 

361.    Theorem.     If  a  line  is  perpendicular  to  a  plane,  any 
jolane  containing  this  line  is  also  perpendicular  to  that  plane. 


Given         OY  perpendicular  to  the  plane  MN,  and  ZY  any 
plane  containing  OY. 

To  prove    that  ZY±  MN. 


Prop.  XIX.]  PENCIL    OF  PLANES.  267 

Proof.    1.  Suppose  OX,  in  MN,  _L  OZ,  the  intersection  of  MN 
and  ZY. 

Then  A  YOZ,  XO  Y  are  right  A.  Why  ? 

2.  But  v  Z  Z07  fixes  the  measure  of  the  dihedral  Z, 

§359 
.'.  Zr_L  JlfiV.  Def. 


Proposition   XIX. 

362.  Theorem.  If  two  planes  are  perpendicular  to  each 
other,  any  line  in  one  of  them,  perpendicular  to  their  inter- 
section,  is  perpendicxdar  to  the  other. 


Given         the  planes  ZYA.MN,    OZ  their  intersection,  and 
OX,  in  MX,  _L  OZ. 

To  prove    that  OX  1.  ZY. 

Proof.    1.  Let  OY,  in  ZY,  he  ±  to  OZ  at  0. 

Then  Z  A^O  Y  is  the  measuring  angle.  §  359 


*  2. 

.-.  A  XOY  is  right. 

Def.  _L  planes 

3.  But 

'  Z  ZOX  is  also  right, 

Why  ? 

.'.  OXA.ZY. 

Why? 

Corollaries. 

1. 

If  two  planes  are  pern 

endicular  to  each 

other,  a  line  from  any  point  in  their  line  of  intersection,  per- 
pendicular to  either,  lies  in  the  other. 

By  the  theorem,   OY  ±  MN,  and  it  lies  in  ZY;  and  by  prop.  VII, 
cor.  2,  only  one  perpendicular  to  MN  can  be  drawn  from  O. 


268 


SOLID    GEOMETRY 


[Bk.  VI. 


2.  Through  a  point  without  a  line  not  more  than  one  plane 
can  pass  perpendicular  to  that  line. 

For  if  through  Y  another  plane  could  pass  _L  OX,  it  would  pass  through 
O,  v  A  XOY  =  rt.  Z,  and  only  one  _L  can  be  drawn  from  Y  to  OX. 
But  the  plane  would  also  include  line  OZ,  else  there  would  be  two  Js 
from  0  to  OX  in  the  plane  MN. 


Proposition   XX. 

363.  Theorem.  If  each  of  two  intersecting  planes  is  per- 
pendicular to  a  third  plane,  their  line  of  intersection  is  also 
perpendicular  to  that  plane. 


Given         two  planes,  Q,  R,  intersecting  in  OP,  and  each  per- 
pendicular to  plane  M. 

To  prove    that  OP  _L  M. 

Proof.    1.   A  _L  to  Jf  from  0  lies  in  Q  and  in  R.  Prop.  XIX,  cor.  1 

2.  .'.  it  coincides  with    OP,  the  only  line  common  to 
Q  and  22.     .\  OP  J_  M. 


Exercises.     564.    To  construct  a  plane  containing  a  given  lino,  and 
parallel  to  another  given  line.     (Assumed  in  step  1  of  prop.  XVII.) 

565.  Prove  that  vertical  dihedral  angles  are  equal. 

566.  How  many  degrees  in  the  measure  of  the  dihedral  angle  between 
the  plane  of  the  earth's  equator  and  the  ecliptic  ? 

567.  Prove  that  the  edge  of  a  dihedral  angle  is  perpendicular  to  the 
plane  of  the  measuring  angle. 

568.  Prove  that  a  line  and  its  projection  on  a  plane  determine  a 
second  plane  perpendicular  to  the  first. 


Prop.  XXI.] 


PENCIL    OB    PLANES. 


269 


Proposition   XXI. 

364.    Theorem.     Any  point   in    a  plane  which    bisects    a 
dihedral  angle  is  equidistant  from  the  faces  of  the  angle. 


Given  a  dihedral  angle,  with  faces  Q,  B,  and  edge  CD, 
bisected  by  plane  B;  P,  any  point  in  B,  with 
PX±  Q,  PY±B. 

To  prove    that  PX=PY. 

Proof.  1.  Let  M  be  the  plane  of  PX,  PF,  and  D  its  intersec- 
tion with  CD. 

2.  Then  vPJlft  .'.  MA.  Q.  Why? 

3.  Similarly,  M _1_  R.  Why? 

4.  .'.MA.  CD.  Prop.  XX 

5.  . ' .  CD  _L  DX,  D  Y,  DP,  whose  A  therefore  measure 
the  dihedral  A.  §  359 

6.  .'.AXDP  =  APDY. 

And  v  A  X=  A  Y=  rt.  Z,  and  DP  =  DP, 

7.  .'.A  DXP  ^ADYP,  and  PX  =  PY.      §  88,  cor.  7 

Corollary.  TAe  focz^s  of  points  that  are  equidistant  from 
two  intersecting  planes  is  the  pair  of  planes  bisecting  their 
dihedral  angles. 


270  SOLID   GEOMETRY.  [Bk.  VI. 

Proposition   XXII. 

365.  Theorem.  If  from  any  point  lines  are  drawn  per- 
pendicular to  two  intersecting  planes,  the  angle  formed  by 
these  perpendiculars  has  a  measure  equal  or  supplemental 
to  that  of  the  dihedral  angle  of  the  planes. 


Given         the  planes  M,  Q,  intersecting  in  i ;  lines  PX  _L  M, 
PY  J_  Q-,  and  plane  PYX  cutting  i  at  A. 

To  prove    that  Z  YPX  is  equal  or  supplemental  to  the  dihe- 
dral angle  MQ. 

Proof.    1.  Plane  YXP  _L  M,  also  _L  Q.  Prop.  XVIII 

2.  .-.  plane  YXP  _L  i.  Prop.  XX 

3.  .'.  XA  and  YA  J_  i.  §  339 

4.  .-.  Z  XA  Y  measures  dihedral  Z  MQ.  §  359 

5.  But  vZZ=Zr=rt.  Z,  Given 
.*.  Z  YPX  is  supplemental  to  AX  AY,  or  dihedral 
Z  Jf©.                                                 I,  prop.  XXI,  cor. 

Corollary.     If  the  point  is  within  the  dihedral  angle,  the 
angles  are  supplemental. 

Definition.     If  two  planes  do  not  meet,  however  far  pro- 
duced, they  are  said  to  be  parallel. 

The  term  "pencil  of  parallels  is  applied  to  planes  as  well  as  to  lines. 


Prop.  XXIII.]  PENCIL    OF  PLANES. 


271 


Proposition   XXIII. 

366.    Theorem.    Planes  perpendicular  to  the  same  straight 
line  are  parallel 


I        x1       M\ 


r~^\ 


Given         two  planes,  M,  X,  J_  line  XY,  at  X,  Y,  respectively. 

To  prove    that  31 II  X. 

Proof.         If  M  and  X  should  meet,  as  at  P,  then  two  planes 
would  pass  through  P  _L  XY,  which  is  impossible. 

Prop.  XIX,  cor.  2 


Exercises.  569.  Prove  that  through  a  point  without  a  plane  any 
number  of  lines  can  pass  parallel  to  the  plane. 

570.  Problem  :  To  bisect  a  dihedral  angle. 

571.  To  find  the  locus  of  points  equidistant  from  two  fixed  planes,  and 
equidistant  from  two  fixed  points. 

I        572.    To  find  a  point  equidistant  from  two  given  planes,  equidistant 
from  two  given  points,  and  also  at  a  given  distance  from  a  third  plane. 

573.  Prove  prop.  XXII  for  the  case  in  which  the  point  P  is  taken  in 
plane  M. 

574.  In  the  figure  on  p.  270,  as  Z  XA  Y  increases  from  zero  to  a 
straight  angle,  what  change  does  Z  YPX  undergo  ? 

575.  Also,  suppose  Z  XAY  =  120°;  what  angle  will  PY  make  with 
plane  M ,  if  produced  through  Q  to  M  ? 

576.  Given  two  points,  F,  W,  in  two  intersecting  planes,  If,  Q, 
respectively.  Find  Z  in  the  line  of  intersection  of  M  and  Q,  such  that 
VZ  +  ZW  shall  be  a  minimum. 

577.  If  from  two  points  on  a  line  parallel  to  a  plane,  parallel  lines  are 
drawn  to  that  plane,  a  parallelogram  is  formed. 


272  SOLID   GEOMETRY.  [Bk.  VI. 

Proposition   XXIV. 

367.    Theorem.     The   lines   in   which   two  p>arallel  planes 
intersect  a  third  plane  are  parallel. 


Given         two  parallel  planes,  M,  N,  intersected  by  a  third 
plane,  T,  in  lines  a,  b. 

To  prove    that  a  II  b. 

Proof.    1.  a  and  b  are  in  the  same  plane  T. 

2.  And  they  cannot  meet,  because  they  are  in  M  and  N 
respectively,  and  M  II  N. 

3.  .'.  they  are  parallel  by  definition. 

Corollary.  A  line  perpendicular  to  one  of  two  parallel 
planes  is  perpendicular  to  the  other. 

Pass  two  planes  through  that  line  and  apply  prop.  XXIV  and  the  def. 
of  a  plane  ±  to  a  line. 

Exercises.  578.  Through  a  given  point  only  one  plane  can  pass 
parallel  to  a  given  plane. 

579.  If  two  parallel  planes  intersect  two  other  parallel  planes,  the 
four  lines  of  intersection  are  parallel. 

580.  Parallel  lines  have  parallel  projections  on  any  plane.  (Suppose, 
as  a  special  case,  that  the  lines  are  perpendicular  to  the  plane.) 

581.  If  two  lines  are  at  right  angles,  are  their  projections  on  any  plane 
also  at  right  angles  ? 

582.  If  two  planes  are  perpendicular  to  each  other,  any  line  perpen- 
dicular to  one  of  them  is  either  parallel  to  or  lies  in  the  other. 


Prop.  XXV.]  PENCIL    OF  PLANES.  273 

Proposition   XXV. 

368.    Theorem.     If  two  straight  lines  are  cut  by  parallel 
planes,  the  corresponding  segments  are  proportional. 


Given         ABC,  DEF,  two  lines  cut  by  planes  P,  Q,  R,  in 
points  A,  B,  C  and  1),  E,  F. 

To  prove    that  AB  :  BC  =  BE :  EF. 

Proof.    1.  Suppose  the  line   GIIF,  drawn  through  F,  II  ABC, 
cutting  P,  Q  at  G,  H,  respectively. 
Then  AC,  GF  determine  a  plane;  also  DF,  GF. 

Prop.  I,  cors.  2,  1 

2.  .-.  AGWBH  II  CF,  and  DG  II  EH.  Why  ? 

3.  .-.  AB  =  GH,  and  BC  =  HF.  I,  prop.  XXIV 

4.  But  v  GH:  HF=DE:  EF,         TV,  prop.  X,  cor.  1 

.'.  AB:BC  =  I>E:  EF.  Subst.  3  in  4 


Exercises.  583.  In  a  gymnasium  swimming  tank  the  water  is  5  ft. 
deep,  and  the  ceiling  is  9  ft.  above  the  water;  a  pole  18  ft.  long  rests 
obliquely  on  the  bottom  of  the  tank  and  touches  the  ceiling.  How  much 
of  the  pole  is  in  the  water  ? 

584.  In  the  figure  of  prop.  XXV,  connect  C  and  D,  and  prove  the 
theorem  without  using  the  line  FG. 


274  SOLID   GEOMETRY.  |Bk.  VI. 


4.     POLYHEDRAL   ANGLES. 

369.  Definitions.  When  a  portion  of  space  is  separated 
from  the  rest  by  three  or  more  planes  which  meet  in  but  one 
point,  the  planes  are  said  to  form,  or  to  include,  a  polyhedral 
angle. 

A  polyhedral  angle  is  also  called  a  solid  angle. 

As  two  intersecting  lines  form  an  infinite  number  of  plane  angles,  but 
the  smallest  is  considered  unless  the  contrary  is 
stated,  and  similarly  with  two  intersecting  planes,  V 

so  three  or  more  intersecting  planes  form  an  infi- 
nite number  of  polyhedral  angles,  but,  as  with 
plane  and  dihedral  angles,  only  the  smallest  is 
considered. 


VAB,    VBC, ,  the 

faces. 


The   lines  of  intersection  of   the  planes 
of  a  polyhedral  angle,  each  with  the  next, 

are  called  the  edges  of  the  polyhedral  angle.  ilP^i^^B 

C 
On  account  of  the  complexity  of  the  general  Apolyhedral  angle. 
figure,  the  planes  which  form  a  polyhedral  angle  v-ABCB.  r,thever- 
are  considered  as  cut  off  by  the  edges,  as  in  the  tex ;  VA,  VB,  VC, 
above  figure.  So  also  the  edges,  which  may  be  ^'J^^jf s;  pla"es 
produced  indefinitely,  are  considered  as  cut  off  by 
the  vertex  unless  the  contrary  is  stated. 

The  portions  of  the  planes  which  form  a  polyhedral  angle, 
limited  by  the  edges,  are  called  the  faces  of  the  angle. 

370.  Polyhedral  angles  contained  by  3,  4, ,  n  planes  are 

termed  respectively  trihedral,  tetrahedral,  /7-hedral  angles. 

A  polyhedral  angle  is  specifically  designated  by  a  letter  at  its  vertex, 
or  by  J,hat  letter  followed  by  a  hyphen,  and  letters  on  the  successive 
edges. 

371.  Congruent  polyhedral  angles  are  such  as  have  their 
dihedral  angles  equal,  and  the  plane  angles  of  their  faces  also 
equal  and  arranged  in  the  same  order. 


Secs.  372,  373.] 


POLYHEDRAL   ANGLES. 


275 


372.  Symmetric  polyhedral  angles  are  such  as  have  their  dihe- 
dral angles  equal,  and  the  plane  angles  of  their  faces  also 
equal,  but  arranged  in  reverse  order. 


A     X        C 

B  B 

Symmetric  polyhedral  angles. 


Opposite  polyhedral  angles. 


Thus,  in  the  above  figure,  V  and  V  are  symmetric  trihedral  angles, 
the  letters  showing  the  reverse  arrangement. 

Some  idea  of  this  reverse  arrangement  may  be  obtained  by  thinking  of 
two  gloves,  fitting  the  right  and  left  hands  respectively.  As  two  such 
gloves  are  not  congruent,  so,  in  general,  two  symmetric  polyhedral  angles 
are  not  congruent. 

373.  Opposite  polyhedral  angles  are  such  that  each  is  formed 
by  producing  the  edges  and  faces  of  the  other  through  the 
vertex. 


Exercises.  585.  How  many  edges  in  an  n-hedral  angle  ?  How  many 
dihedral  angles  ?     How  many  plane  face  angles  ?     How  many  vertices  ? 

586.  If  a  straight  line  is  oblique  to  one  of  two  parallel  planes,  it  is  to 
the  other. 

587.  If  a  plane  intersects  all  the  faces  of  a  tetrahedral  angle,  what 
kind  of  a  plane  figure  is  formed  by  the  lines  of  intersection  ?  What,  in 
the  case  of  a  trihedral  angle  ? 

588.  Does  the  magnitude  of  a  polyhedral  angle  depend  upon  the 
lengths  of  the  edges  ?  « 

589.  Construct  from  stiff  paper  two  symmetric  trihedral  angles,  with 
face  angles  of  about  30°,  60°,  45°,  and  see  if  they  are  congruent.  (No 
proof  required.) 

590.  If  each  of  two  intersecting  lines  is  parallel  to  a  plane,  so  is  the 
plane  of  those  lines. 


276 


SOLID   GEOMETRY. 


TBk.  VI. 


Proposition   XXVI. 
374.    Theorem.     Opposite  polyhedral  angles  are  symmetric. 
B' 


Given  V-ABCD,  any  polyhedral  angle,  and  V-A'B'C'D',  its 

opposite  polyhedral  angle. 

To  prove    that  V-ABCD  and  V-A'B'C'D'  are  symmetric. 
Proof.    1.  Z.AVB  =  AA'VB', 

ZBVC  =  ZB'  VC, Prel.  prop.  VI 

2.  Dihedral  A  with  edges  VB,  VB',  being  formed  by 
the  same  planes,  have  equal  (vertical)  measuring 
angles.  Prel.  prop.  VI 

3.  So  for  the  other  dihedral  A.  But  the  order  of 
arrangement  in  the  one  is  reversed  in  the  other. 
.*.  the  polyhedral  A  are  symmetric. 

Note.  That  the  order  of  the  angles  is  reversed  appears  more  clearly 
to  the  eye  by  making  two  opposite  trihedral  angles  of  pasteboard.  It  is 
also  seen  by  tipping  the  upper  angle  over,  as  has  been  done  in  the  figure 
to  the  right. 

Exercises.  591.  If  the  edges  of  one  polyhedral  angle  are  respectively 
perpendicular  to  the  faces  of  a  second  polyhedral  angle,  then  the  edges  of 
the  latter  are  respectively  perpendicular  to  the  faces  of  the  former. 

592.  Two  parallel  planes  intersecting  two  parallel  lines  cut  off  equal 
segments. 


Prop.  XXVII.] 


POLYHEDRAL   ANGLES. 


277 


Proposition   XXYII. 

375.    Theorem.     In  any  trihedral    angle  the  sum  of  any 
two  face-angles  is  greater  than  the  third. 


Given         the  trihedral  angle  V-XYZ. 

To  prove    that  ZLYVZ  +  ZZ  VX  >  Z  XV Y. 

Proof.    1.   If  Z  XVY>  either  Z  YVZ  or  Z  ZVX,  no  proof  is 
necessary.     Why  not  ? 

2.  If  ZA7T  >  either  Z  FFZ  or  Z  ZVX,  suppose  it 
>  A  ZVX. 

3.  Then  in  plane    VXY  suppose    VW  drawn,  making 
Z  XVW  =  Z  ZVX. 

Suppose  VC  taken  on  VZ,  equal  to  VP  on  VW,  and 
a  plane  passed  through  C,  P,  and  any  point  A  of 
FX.     Let  this  plane  intersect  VY  at  P. 

4.  Then  A  ^  FP  ^  A  ^  TC,  and  A  C  =  AP.     I,  prop.  I 

5.  But  v  AC+CB>  AB,  or  AP  +  PB,  Why  ? 

.-.  CP  >  PB.  Why? 

6.  .'.  in  APF^and  CVB,  ZBVO  ZPVB. 

I,  prop.  XI 

7.  . • .  Z  C  FJ  +  Z  £  JrC>  Z A  VP  +  Z  P VB,  or  Z  yl  FP. 
Or  Z  TFZ  -f-  Z  Z  VX  >  Z  IFF.  Ax.  4 


278  SOLID   GEOMETRY.  [Bk.  VI. 

Corollaries.  1.  In  any  trihedral  angle  the  difference  of 
any  two  face-angles  is  less  than  the  third. 

For  if  the  face-angles  are  a,  6,  c,  then  since  a  +  6  > c,  .-.a>c  —  b. 

2.  In  any  'polyhedral  angle  any  face-angle  is  less  than  the 
sum  of  all  the  other  face-angles. 

For  the  polyhedral  angle  may  be  divided  into  a  number  of  trihedral 
angles,  and  prop.  XXVII  repeatedly  applied. 

Note.  Prop.  XXVII  and  corollaries  suppose  that  each  face-angle  is 
less  than  a  straight  angle.  This  is  in  accordance  with  the  note  under 
the  definition  of  a  polyhedral  angle. 

376.  Definition.  A  polyhedral  angle  is  said  to  be  convex 
when  any  polygon  formed  by  a  plane  cutting  every  face,  is 
convex ;  otherwise  it  is  said  to  be  concave. 


Proposition  XXVIII. 


p* 


377.    Theorem.     In  any  convex  polyhedral  angle  the  sum 
of  the  face-angles  is  less  than  a  perigon. 


Given         any  convex  polyhedral  angle,  V-ABC 

To  prove    that  Z  A  VB  +  Z  B  VC  +  Z  C  VI)  + <  perigon. 

Proof.  1.  Let  the  faces  of  the  angle  be  cut  by  a  plane.  This 
will  form  a  convex  polygon  of  n  sides  (n  =  5  in  the 
figure),  abc Def.  convex  polyh.  Z 


Prop.  XXV1II.J  POLYHEDRAL   ANGLES.  279 

Let     Sv  =  sum  of  plane  A  a  Vb,  bVc,  ,   at  the 

vertex ; 
Sb  =  sum   of  plane  AbaV,    Vba,  cbV, ,  at 

the  bases  of  the  A  ; 
and     ^  =  sum    of    plane   A  cba,    deb,  ,    of    the 

polygon. 

2.  Then  Sp  =  (n  -  2)  st.  A,  or  Sp  +  2  st.  A  =  n  st.  A. 

I,  prop.  XXI 

3.  And  Sv  +  Sb  =  n  st.  A,  since  there  is  a  st.  Z  for 

each  A.  I,  prop.  XIX 

4.  .".  8V  +  Sb  =  ^  +  perigon.         Steps  2  and  3  ;  ax.  1 
5.  And  V  Sb  >  Sp,  Prop.  XXVII 

.'.  Sv  <  perigon. 


Exercises.  593.  The  three  planes  which  bisect  the  three  dihedral 
angles  of  a  trihedral  angle  intersect  in  a  common  line  whose  points  are 
equidistant  from  the  three  faces.  (See  prop.  XXI,  cor.,  and  I,  prop. 
XLIV.) 

594.  Suppose  a  polyhedral  angle  formed  by  three,  four,  five  equilateral 
triangles.     What  is  the  sum  of  the  face-angles  at  the  vertex  ? 

595.  If  lines  through  any  point  0  and  the  vertices,  A,  B,  C,  ,  of 

a  polygon,  cut  a  plane  parallel  to  the  plane  of  that  polygon  in  A',  B',  C", 

,  prove  that  A'B'C *>~  ABC and  that  the  ratio  of  similitude 

is  that  of  OA'  to  OA. 

596.  In  ex.  595,  the  more  remote  0  is  from  the  planes  ABC , 

A'B'C' ,  the  more  nearly  do  AA',  BB',  CC become  parallel; 

suppose  they  become  parallel,  state  and  prove  the  resulting  theorem. 

597.  In  ex.  595,  if  plane  A'B'C were  not  parallel  to  plane  ABC 

,  prove  that  the  corresponding  sides,  AB,  A'B',  and  BC,  B'C,  and 

CD,  CD',  ,  would,  in  general,  meet  in  points  on  the  intersection  of 

the  two  planes. 

598.  Two  planes,  each  parallel  to  a  third  plane,  are  parallel  to  each 
other. 

599.  Ex.  598  is  analogous  to  I,  prop.  XVIII.  State  the  theorem  and 
corollaries  analogous  to  I,  prop.  XVII  and  its  corollaries,  and  prove  them. 


280 


SOLID    GEOMETRY. 


[Bk.  VI. 


5.     PROBLEMS. 
Proposition  XXIX. 

378.  Problem.  Through  a  given  point  to  pass  a  plane 
perpendicular  to  a  given  line :  (1)  the  point  being  without 
the  line,  (2)  the  point  being  on  the  line. 


1.  Given    the  line  YY',  and  point  P  without. 
Required    through  P  to  pass  a  plane  _L  YY'. 

Construction.    1.  From  P  draw  PO  _L  YY'.  I,  prop.  XXX 

2.  From  0  draw  another  line  OX  A.  YY'. 

I,  prop.  XXIX 
Then  MN,  the  plane  of  OP,  OX,  is  the  required 
plane. 

Proof.  V  YY'±OP, 

and  YY'  _L  OX, 

.'.  YY'±MX.  §339 

2.  Given    the  line  YY',  and  the  point  0  upon  it. 
Required    through  0  to  pass  a  plane  _L  IT. 

Construction  and  Proof.     Draw  OP  and  0X1.  YY'.     This  can 
be  done  because  the  three  lines  are  not  required  to 
be  coplanar. 
Then  the  plane  XOP  1  V  V.  §  339 


Props.  XXX,  XXXL] 


PROBLEMS. 


281 


Proposition  XXX. 

379.  Problem.  Through  a  given  point  to  pass  a  plane 
parallel  to  a  given  plane. 

Solution.  Draw  two  intersecting  lines  in  the  given  plane. 
Through  the  given  point  draw  two  lines  parallel  to  these  lines, 
thus  determining  the  required  plane. 


Proposition  XXXI. 

380.  Problem.  Through  a  given  point  to  draw  a  line  per- 
pendicular to  a  given  plane  :  (1)  the  point  being  without  the 
plane,  (2)  the  point  being  in  the  plane. 


1.  Given    the  plane  MN,  and  the  point  P  without. 
Required    to  draw  a  perpendicular  from  P  to  MN. 

Construction.    1.  Draw  PC  ±AB,  any  line  in  MN. 

I,  prop.  XXX 
2.  In  MN  draw  CE  _L  AB.  I,  prop.  XXIX 


Draw 


PP'  _L  CE. 


I,  prop.  XXX 


Then  PP'  is  the  required  perpendicular. 
Proof.    1.  CA  _L  plane  CPP'.  Prop.  VI,  cor.  1 

2.  Draw  P'D  II  CA  ;  then  P'D  _L  plane  CPP'.    Prop:  IX 

3.  .-.  Z  DP'P  is  right,  and  PP'  X  P'D.  §  339 

4.  But  PP'  _L  CP'.  and  .'.  PP'  ±  MX.   Prop.  VI,  cor.  1 


282  SOLID   GEOMETRY. 

2.  Given    the  plane  MN,  and  the  point  R  within  it. 


[Bk.  VI. 


Required    through  R  to  draw  a  perpendicular  to  MN. 

Construction.    1.  From  any  external  point  S  draw  ST  JL  MN. 

Case  1 
2.  From  R  draw  RQ  II  TS.  I,  prop.  XXXIII 

Proof.         Then  RQ  is  the  required  perpendicular.  Why  ? 


Exercises.     600.    From  the  point  of  intersection  of  two  lines  to  draw 
a  line  perpendicular  to  each  of  them. 
>      601.   To  determine  the  point  whose  distances  from  the  three  faces  of 
a  given  trihedral  angle  are  given.     Is  it  unique  ? 

602.    From  the  vertex  of  a  trihedral  angle  to  draw  a  line  making  equal 
angles  with  the  three  edges. 
f        603.    The  three  planes,  through  the  bisectors  of  the  face-angles  of  a 
trihedral  angle,  perpendicular  to  those  faces,  intersect  in  a  common  line 
wdiose  points  are  equidistant  from  the  edges.     (See  I,  prop.  XLIII.) 

604.  In  how  many  ways  can  a  polyhedral  angle  be  formed  with  equi- 
lateral triangles  and  squares  ? 

605.  Prove  that  a  straight  line  makes  equal  angles  with  parallel  planes. 

606.  If  each  of  two  intersecting  planes  is  parallel  to  a  given  line,  prove 
that  their  intersection  is  coplanar  with  that  line. 

607.  Frove  that  parallel  lines  make  equal  angles  with  parallel  planes. 

608.  Are  planes  perpendicular  to  the  same  plane  parallel  ? 

609.  In  the  figure  of  prop.  XXV,  without  drawing  FG,  draw  CD  and 
AF;  then  show  that  the  four  lines  CD,  CA,  FD,  FA  intersect  plane  Q 
in  the  vertices  of  a  parallelogram. 

610.  Given  two  lines,  not  coplanar,  and  a  plane  not  containing  either 
line,  required  to  draw  a  straight  line  which  shall  cut  both  given  lines  and 
shall  be  perpendicular  to  the  plane.     (Project  both  lines  on  the  plane.) 


BOOK  VII.  —  POLWEDBA. 


1.  GENERAL  AND  REGULAR  POLYHEDRA. 

381.  Definitions.  A  solid  whose  bounding  surface  consists 
entirely  of  planes  is  called  a  polyhedron;  the  polygons  which 
bound  it  are  called  its  faces ;  the  sides  of  those  polygons,  its 
edges ;  and  the  points  where  the  edges  meet,  its  vertices. 

382.  If  a  polyhedron  is  such  that  no  straight  line  can  be 
drawn  to  cut  its  surface  more  than  twice,  it  is  said  to  be 
convex;    otherwise  it  is  said  to  be  concave. 

Unless  the  contrary  is  stated,  the  word  polyhedron  means  convex  poly- 
hedron. The  word  convex  will,  however,  be  used  wherever  necessary  for 
special  emphasis. 

383.  If  the  faces  of  a  polyhedron  are  congruent  and  regular 
polygons,  and  the  polyhedral  angles  are  all  congruent,  the 
polyhedron  is  said  to  be  regular. 


Exercises.  611.  Draw  a  figure  of  a  polyhedron  of  four  faces.  Count 
the  edges,  faces,  and  vertices,  and  show  that  the  number  of  edges  plus 
two  equals  the  number  of  faces  plus  the  number  of  vertices. 

612.  Do  the  same  for  a  polyhedron  of  five  faces  ;  also  of  six  faces. 

613.  Take  a  piece  of  chalk,  apple,  or  potato,  and  see  if  a  seven-edged 
polyhedron  can  be  cut  from  it. 

614.  What  is  the  locus  of  points  on  the  surface  of  a  polyhedron  equi- 
distant from  two  given  vertices  ?  (The  distances  are  to  be  taken  as  usual 
on  a  straight  line,  and  not  necessarily  on  the  surface.) 

615.  What  is  the  locus  of  points  equidistant  from  two  given  non- 
parallel  faces  of  a  given  polyhedron  ? 

616.  To  find  a  point  equidistant  from  two  given  vertices  of  a  polyhe- 
dron, and  from  two  given  non-parallel  faces. 

283 


284 


SOLID   GEOMETRY 


[Bk.  VII. 


Proposition  I. 

384.    Theorem.     If  a  convex  polyhedron  has  e  edges,  v  ver- 
tices, and  f  faces,  then  e  +  2  =  f  +  v. 


Given 

To  prove 
Proof.    1 


ABC Z,  a  convex  polyhedron  of  e  edges,  v  ver- 
tices, /  faces. 

that  e  +  2=f+v. 

Imagine  ABC Z  formed  by  adding  adjacent  faces, 

beginning  with  any  face  as  ABCD of  a  sides, 

then  adding  face  M,  of  b  sides,  and  so  on. 

(It  is  advisable  to  build  up  a  rectangular  box  of  paste-board 
while  reading  the  proof.) 

Let  er  =  the  number  of  edges,  and  vr  =  the  number 
of  vertices,  after  r  faces  have  been  put  together. 

(Thus  when  we  put  2  rectangles  together  in  building  up  the 
box,  we  have  located  7  edges  and  6  vertices;  i.e.  e»  =  7, 
v2  =  C,  in  this  case.) 

2.  Then,  since  the  first  face  had  a  sides,  .*.  ex  =  a  and 
vl  =  a. 

(In  the  box,  ex  =  4,  vi  =  4.) 

3.  v  adding  an  adjacent  face  71/,  of  b  sides,  gives  only 
(b  —  1)  new  edges,  and  (b  —  2)  new  vertices  (Why  ?), 

(In  the  box,  adding  a  second  rectangle  to  the  first  gives 
only  3  new  edges  because  we  have  1  in  common  with  the 
first  face,  and  2  new  vertices  because  we  have  2  in  common 
with  the  first  face.) 

.'.  e2  =  a  -f-  b  —  1,  v2  =  a  +  b  —  2,  so  that  ea  —  v2  =  1. 
(In  the  box,  e2  =  4  +  4  -  1  =  7,   v2  =  4  +  4  -  2  =  C,  so 
that  e2  —  «a  =  1.) 


Prop.  I.]    GENERAL   AND   REGULAR   POLYIIEDRA.  285 

4.  Therefore  we  have 

e2  =  *>a  +  1. 

Now  while  the  number  of  edges  common  to  two 
successive  open  surfaces  will  vary  according  to  the 
way  in  which  the  additions  are  made,  the  addition 
of  a  new  face  evidently  increases  e  by  one  more 
unit  than  it  increases   v. 

.-.es^Vs  +  2, 
€i  =  r4  +  3, 
and,  in  general, 

er  =  vr  +  r  -  1, 
or  er  —  vr  =  r  —  1. 

5.  But  the  addition  of  the  last,  or  /th  face,  as  XYZ, 
after  all  the  others  have  been  put  together,  gives  no 
new  edges  or  vertices, 

•'•  ef -  l'f  =  e/-i  -  vf-i  =f~2' 
(In  the  box,  adding  the  last  face  merely  puts  on  the  cover, 
adding  no  new  edges  or  vertices.     .-.  e6  —  »e  =  e5  —  v5  =  4, 
which  is  evidently  true,  because  12,  the  number  of  edges, 
minus  8,  the  number  of  vertices,  is  4.) 

6.  That  is,  e  -  v  =f-  2,  so  that  e  +  2  =/+  u;   for 

ef  =  e,  and  vf  =  v. 

Corollary.  For  every  polyhedron  there  is  another  which, 
with  the  8<i.me  number  of  edges,  has  as  many  faces  as  the  first 
has  vertices,  and  as  many  vertices  as  the  first  has  faces. 

It  is  easily  seen  that  a  polyhedron  can  be  inscribed  with  a  vertex  at  the 
center  of  each  face,  the  number  of  edges  remaining  the  same. 

Note.  This  theorem  is  known  as  Euler's,  although  Descartes  knew 
and  employed  it.     The  theorem  is  very  useful  in  the  study  of  crystals. 


Exercise.  617.  If  the  faces  of  a  polyhedron  are  all  triangular,  the 
number  of  faces  is  even  and  is  four  less  than  twice  the  number  of  ver- 
tices. (Since  there  are  3  edges  to  every  face,  but  each  edge  belongs  to 
2  adjacent  faces,  e  =  3//2 ;  substitute  in  e  +  2  =/  +  v.) 


I 


286  SOLID   GEOMETRY.  [Bk.  VII. 

Proposition  II. 

385.  Theorem.  There  cannot  be  more  than  jive  regular 
convex  polyhedra. 

Proof.  1.  Let  n  =  number  of  sides  in  one  face,  and  a  =  number 
of  degrees  in  each  plane  Z  of  the  faces  of  a  regular 
convex  polyhedron. 

Then  a  =  (n  -  2)  •  180°/ n,  I,  prop.  XXI 

and  if  n  =  6,  then  a  =  120°,  and  3  a  =  360°. 

2.  .'.  if  n  =  6  or  more,  there  can  be  no  solid  angle. 

VI,  prop.  XXVIII 

3.  And  if  n  =  5,  then  a  =  108°,  and  3a  =  324°, 

and  .'.3  regular  pentagons,  but  no  more,  can  form  a 
solid  angle.  VI,  prop.  XXVIII 

4.  And  if  n  =  4,  then  a  =  90°,  3  a  =  270°,  4  «  =  360°, 
and  .'.3  squares,  but  no  more,  can  form  a  solid 
angle.  VI,  prop.  XXVIII 

5.  And  if  n  =  S,  a  =  60°,  3  a  =  180°,  4  a  =  240°, 
5  a  =  300°,  6a  =  360°, 

and  .*.  3,  4,  or  5  equilateral  A,  but  no  more,  can 
form  a  solid  angle:  VI,  prop.  XXVIII 

6.  .'.  there  cannot  be  more  than  5  regular  -convex 
polyhedra,  viz.  those  formed  by  regular  pentagons 
(3  at  each  vertex),  squares  (3  at  each  vertex),  equi- 
lateral triangles  (3,  4,  or  5  at  each  vertex). - 

Note.  There  are  five  regular  convex  polyhedra ;  but  the  complete 
proof  of  the  fact  is  not  of  enough  importance  to  insert  it  in  the  body  of 
the  work.  It  may  be  given  as  an  exercise,  since  it  involves  no  new 
principles.  These  five  polyhedra  have  been  called  the  Platonic  Bodies, 
from  the  attention  given  them  in  Plato's  school,  although  they  were 
known  to  the  Pythagoreans.  The  three  simpler  forms  enter  largely  into 
crystallography,  usually  somewhat  modified. 

The  five  regular  polyhedra  are  given  on  page  287. 


Prop.  II.]       GENERAL   AND  REGULAR   POLYHEDRA.         287 


The  regular  tetrahedron 
(or  triangular  pyramid), 
formed  by  4  equilateral 
triangles. 


The  regular  hexahedron 
(or  cube),  formed  by 
6  squares. 


The  regular  octahedron, 
formed  by  8  equilateral 
triangles. 


The  regular  dodecahedron,  formed  by 
12  regular  pentagons. 


The  regular  icosahedron,  formed  by 
20  equilateral  triangles. 


The  five  regular  polyhedra  can  be  constructed  from  cardboard  by- 
marking  out  the  following,  cutting  through  the  heavy  lines  and  half 
through  the  dotted  ones,  and  then  bringing  the  edges  together. 


Tetrahedron. 


Hexahedron. 


Octahedron. 


Dodecahedron. 


Icosahedron. 


288  SOLID    GEOMETRY.  [Bk.  VII. 


2.     PARALLELEPIPEDS. 

386.  Definitions.  A  Parallelepiped  is  a  solid  bounded  by- 
three  pairs  of  parallel  planes. 

The  four  lines  through  a  parallelepiped,  joining  the  oppo- 
site vertices,  are  called  its  diagonals. 


Proposition   III. 

387.  Theorem.  The  opposite  faces  of  a  parallelepiped  are 
congruent  parallelograms ;  and  any  section  of  it,  made  hrj 
a  plane  cutting  two  pairs  of  opposite  faces  without  cutting 
the  remaining  pair,  is  a  parallelogram. 


Given         the  parallelepiped  AG,  and  PQfiS  a  plane  section 
cutting  the  parallel  faces  AF,  DG,  and  AH,  BG. 

To  prove    (1)  that  A  C  and  KG   are  congruent  UJ, 
(2)  that  PQHS  is  a  O. 

Proof.    1.  EF  II  HG  II  DC  II  AB  II  EF, 

and  BC  II  FG  II  EH  II  AD  II  BC,   VI,  prop.  XXIV 

and  .'.all  faces  are  UJ.               §  (.)7,  del  O 

2.  . ■ .  AB  =  EF  =  JIG  =  D C, 

and  BC  =  FG  =  Ell  =  A  J).       I,  prop.  XXIV 


Prop.  HI.]  PARALLELEPIPEDS.  289 

3.  And  v  Z  JEKff  =  Z  7/  J  Z>,  VI,  prop.  V 

.-.  O  AC ^  CD  EG,  which  proves  (1).    I,  prop.  XXVI 
Similarly  for  other  opposite  faces. 

4.  And   v  PQ  II  SB,  and  P,S'  II  QE,      XI,  prop.  XXIV 

.-.  Pi?  is  a  O,  which  proves  (2).     §  97.  def.  O 

Corollary.  A  parallelepiped  has  three  sets  of  pta.rallel 
and  equal  edges,  four  in  each  set. 

388.  Definition.  If  the  faces  of  a  parallelepiped  are  all 
rectangles,  it  is  called  a  rectangular  parallelepiped. 

It  will  be  noticed  that  as  axes  of  symmetry  enter  into  the  study  of 
plane  figures  (§  68),  and  especially  of  regular  figures,  so  planes  of  sym- 
metry and  axes  of  symmetry  enter  into  the  study  of  solids.  A  plane  of 
symmetry  divides  the  solid  into  halves,  related  to  each  other  as  a  figure  is 
related  to  its  image  in  a  mirror.  Planes  of  symmetry  play  an  important 
part  in  the  study  of  crystals.  The  term  axis  of  symmetry  will  be  under- 
stood from  Plane  Geometiy. 


Exercises.  618.  Prepare  a  table  showing  the  number  (1)  of  faces, 
(2)  of  edges,  (3)  of  vertices,  (4)  of  sides  in  each  face,  (5)  of  plane  angles 
at  each  vertex,  of  all  of  the  five  regular  polyhedra. 

619.  How  many  degrees  in  the  sum  of  the  face-angles  at  one  vertex 
of  a  regular  tetrahedron?  hexahedron?  octahedron?  dodecahedron? 
icosahedron  ? 

620.  The  perpendiculars  to  the  faces,   through  their  centers,   of  a       ^ 
regular  tetrahedron  are  concurrent  in  n  point  equidistant  from  all  of  the 
vertices,  from  all  of  the  faces,  and  from  all  of  the  edges. 

621.  Prove  that  no  polyhedron  can  have  less  than  six  edges. 

622.  In  a  regular  tetrahedron  three  times  the  square  on  an  altitude    -W* 
equals  twice  the  square  on  an  edge. 

623.  Certain  crystals  have  their  corners  cut  off,  that  is,  the  vertices  of 
their  polyhedral  angles  replaced  by  planes.  Suppose  a  regular  hexa- 
hedral  crystal  has  its  trihedral  angles  replaced  by  planes,  how  many  faces 
has  the  new  crystal  ?  How  many  edges  ?  vertices  ?  Is  Euler's  theorem 
satisfied  ? 

624.  How  many  planes  of  symmetry  and  how  many  axes  of  sym- 
metry has  a  regular  hexahedron  ?   octahedron  ? 


f 


290 


SOLID   GEOMETRY. 


[Bk.  VII. 


Pkopositiox  IV. 

389.  Theorem.     In  any  parallelepiped, 

1.  The  four  diagonals  are  concurrent  in  the  mid-point 
of  each. 

2.  The  sum  of  the  squares  on  the  four  diagonals  equals 
the  sum  of  the  squares  on  the  twelve  edges. 


A  B 

Given         a  parallelepiped  with  diagonals  AG,  BH,  CE,  DF. 

To  prove    that  (1)  the  diagonals  are  concurrent  at  0,  the  mid- 
point of  each ; 
(2)  AG2  +  BH2  +  CE2  +  BE2  =  AB2  +  BC2  + 

Proof.    1.  v  BF=  and  II  BR,  Why  ? 

.-.DBFH  is  a.  CJ.  Why? 

2.  .'.  ED  and  BH  bisect  each  other  at  0.  §  100,  cor.  2 
Similarly,  BH  and  CE,  CE  and  A  G,  bisect  each  other. 

3.  And  '.'  there  is  only  one  point  of  bisection  of  BH 
and  CE,  §  41 
.  * .  BH,  CE,  A  G,  and  DF  are  concurrent  at  0. 

4.  And  v  AG2  +  CE2  =  AC2  +  CG2  -f-  GE2  +  EA*} 
and  DF2  +  BH2  =  BF2  +  FH2  h  HB2  +  BB2, 

II,  prop.  XI,  cor. 

5.  .'.  by  adding,  and  noting  that  AC2  +  7>/>>2  =  AB2  + 
BC2  +  CD2  +  i>vi'2,  etc.,  the  theorem  is  proved. 


Secs.  390-393.]     PRISMATIC  AND  PYRAMIDAL   SPACE.       291 


3.     PRISMATIC  AND   PYRAMIDAL   SPACE. 
PRISMS   AND   PYRAMIDS. 

390.  Definitions.  A  prismatic  surface  is  a  surface  made  up 
of  portions  of  planes,  the  intersections  of  which  are  all  parallel 
to  one  another. 


391.  If,  counting  from  any  plane  of  a  prismatic  surface 
as  the  first,  each  plane 
intersects  its  succeeding 
plane,  and  the  last  one 
intersects  the  first,  the  sur- 
face is  said  to  enclose  a 
prismatic  space. 

The  lines  of  intersec- 
tion are  called  the  edges, 
and  the  portions  of  the 
planes  between  the  edges, 
the  faces,  of  the  prismatic  space 


A  prismatic  sur- 
face. 


A  portion  of  a  prismatic 
space,  quadrangular 
and  convex.  ABCD, 
a  right  section. 


The  edges  and  the  faces  are  supposed  to  be  unlimited  in  length.  It 
will  be  readily  seen  that  a  prismatic  space  is  related  to  entire  space  as  a 
plane  polygon  is  to  its  entire  plane.  It  will  therefore  be  inferred  that 
theorems  relating  to  polygons  have  corresponding  theorems  relating  to 
prismatic  spaces. 

392.  A  section  of  a  prismatic  space,  made  by  a  plane  cutting 
its  edges,  is  called  a  transverse  section.  If  it  is  perpendicular 
to  the  edges,  it  is  called  a  right  section. 

393.  A  prismatic  space  is  said  to  be  triangular,  quadrangular, 

rectangular,  pentagonal,  ,  n-gonal,  according  as  a  transverse 

section  is  a  triangle,  quadrilateral,  rectangle,  pentagon,  , 

n-gon,  and  to  be  convex  or  concave  according  as  a  transverse 
section  is  a  convex  or  a  concave  polygon. 


292  SOLID   GEOMETRY.  [Bk.  VII. 

Prismatic  spaces  may  be  such  that  transverse  sections  are  convex,  con- 
cave, or  cross  polygons.  All  theorems  not  involving  mensuration  will 
at  once  be  seen  to  apply  to  each  class.  But  on  account  of  the  complexity 
of  the  figures,  the  third  form  (cross)  is  not  considered  in  this  work. 

394.  The  portion  of  a  prismatic  space  included  between  twro 
parallel  transverse  sections  is  called  a  prism,  the  two  transverse 
sections  being  called  the  bases  of  the  prism. 

Thus  in  the  figure  on  p.  293  the  portion  of  the  prismatic  space  P, 
between  S  and  S',  is  a  prism.     S  and  S'  are  the  basis. 

The  signification  of  the  terms  edges,  faces,  and  prismatic  surface  of  a 
prism,  upper  and  lower  bases  of  a  prism,  triangular  prisms,  etc.,  will  be 
inferred  from  the  above  definitions.  By  transverse  and  right  sections  of 
a  prism  are  to  be  understood  the  transverse  and  right  sections  of  its 
prismatic  space. 

The  sides  of  the  bases  of  a  prism  are  also  called  edges ;  where  con- 
fusion is  liable  to  arise  these  are  called  base  edges,  and  the  edges  of  the 
prismatic  space  are  called  lateral  edges. 

:# 

Exercises.  625.  In  the  figure  of  prop.  IV,  prove  that  Oi,  0,  02  are 
collinear. 

626.  Also  that  0X0  =  EA/2. 

627.  Also  that  if  AG  is  a  rectangular  parallelepiped,  0\0  is  perpen- 
dicular to  line  EG. 

628.  Also  that  if  the  diagonals  of  all  the  faces  are  drawn,  and  the 
points  of  intersection  of  the  diagonals  of  the  opposite  faces  are  con- 
nected, these  connecting  lines  are  concurrent  at  O,  the  mid-point  of  each. 

629.  Prove  that  the  square  on  a  diagonal  of  a  rectangular  parallele- 
piped equals  the  sum  of  the  squares  on  three  concurrent  edges. 

630.  If  the  edge  of  a  cube  is  represented  by  Vs,  find  the  diagonal. 

631.  Prove  that  the  four  diagonals  of  a  rectangular  parallelepiped 
are  equal. 

632.  Show  that  the  edge,  diagonal  of  a  face,  and  diagonal,  of  a  cube, 
are  proportional  to  1,  V2,  Vij. 

633.  If  the  plane  PR,  of  prop.  Ill,  were  also  to  cut  the  faces  UF 
and  1)1>,  what  would  be  the  plane  figure  resulting?  What  would  be  the 
relation  of  its  opposite  sides? 


Prop.  Y.]      PRISMATIC   AND  PYRAMIDAL   SPACE.  293 

Proposition  V. 

395.    Theorem.     Parallel  transverse  sections  of  a  prismatic 
space  are  congruent  polygons. 


Given         the  prismatic  space  P,  with  S,  S',  two  parallel  trans- 
verse sections. 

To  prove    that  S  ^  S'. 

Proof.    1.  '•'  the  sides  of  S  II  sides  of  S',  respectively, 

VI,  prop.  XXIV 
.\  A  of  £  =  A  of  S',  respectively.  VI,  prop.  V 

2.  And  '•*  the  sides  of  >S'  also  equal  the  sides  of  S', 
respectively,  I,  prop.  XXIV 

.'.  by  superposition,  S  is  evidently  congruent  to  S'. 

Corollaries.     1.   The  bases  of  a  prism  are  congruent  poly- 
gons. 

2.  The  faces  of  a  j^rism  are  parallelograms. 

3.  The  lateral  edges  of  a  prism  are  equal. 


Exercise.  634.  Suppose  in  the  figure  of  prop.  Ill,  another  plane  II  to 
PR,  cutting  the  same  faces  as  PR,  hut  not  the  other  faces.  Prove  that 
it  would  cut  out  a  parallelogram  congruent  to  PR. 


294  SOLID   GEOMETRY.  [Bk.  VII. 

396.    Definitions.     A  pyramidal  surface  is  a  surface  made  up 
of  portions  of  planes  which  have  but  one  point  in  common. 


A  pyramidal  surface.  A  portion  of  a  pyramidal  space,  quad- 

rangular and  convex.  V,  the  vertex; 
ABCD,  a  transverse  section;  V-ABCD, 
a  pyramid,  ABCD  being  its  base. 

397.  If,  counting  from  any  plane  of  a  pyramidal  surface  as 
the  first,  each  plane  intersects  its  succeeding  plane,  and  the 
last  one  intersects  the  first,  the  surface  is  said  to  contain  a 
pyramidal  space. 

Unlike  a  prismatic  space,  a  pyramidal  space  is  double,  its  parts  lying 
on  opposite  sides  of  the  common  point. 

The  lines  of  intersection  of  the  planes  are  called  the  edges, 
the  portions  of  the  planes  between  the  edges,  the  faces,  and 
the  point  of  intersection  of  the  edges,  the  vertex,  of  the 
pyramidal  space. 

The  edges  and  faces  are  supposed  to  be  unlimited  in  length. 

398.  A  section  of  a  pyramidal  space,  made  by  a  plane 
cutting  all  of  its  edges  on  the  same  side  of  the  vertex,  is  called 
a  transverse  section. 

399.  The   terms   triangular,  ,    n-gonal,    concave,    convex 

pyramidal  space  are  defined  as  the  like  terms  for  prismatic 
space. 


Secs.  400,  401.]     PRISMATIC  AND  PYRAMIDAL   SPACE.      295 


400.  The  portion  of  a  pyramidal  space  included  between 
the  vertex  and  a  transverse  section  is  called  a  pyramid,  the 
transverse  section  being  called  its  base,  and  the  vertex  of  the 
space,  the  vertex  of  the  pyramid. 

Thus  the  figure  V-ABC  below  is  a  pyramid,  ABC  being  the  base  and 
V  the  vertex. 

The  distance  from  the  vertex  of  a  pyramid  to  the  plane  of 
its  base  is  called  the  altitude  of  the  pyramid. 

Thus  in  the  figure  below,  YV  is  the  altitude  of  pyramid  V-ABC. 
The  signification  of  the  terms  edges,  faces,  transverse  section,  base  edges, 
etc. ,  of  a  pyramid  can  be  inferred  from  the  preceding  definitions. 

401.  The  portion  of  a  pyramidal  space  included  between 
two  transverse  sections  on  the  same  side  of  the  vertex  is 
called  a  truncated  pyramid ;  if  the  transverse  sections  are 
parallel,  it  is  called  a  frustum  of  a  pyramid,  the  two  sections 
being  called  the  bases  of  the  frustum. 

A  frustum  of  a  pyramid  is  therefore  a  special  form  of  a  truncated 
pyramid. 

A  pyramid  is  also  a  special  case  of  a  truncated  pyramid,  the  upper 
base  being  zero. 

The  distance  from  any  point  in  one  base  of  a  frustum  of  a 
pyramid  to  the  plane  of  the  other  base  is  called  the  altitude 
of  the  frustum. 


T,  a  truncated  pyramid  ;  ABCXYZ,  a  frustum  of  the  pyramid  V-ABC ; 
VV,  the  altitude  of  the  pyramid  ;  ABC,  XYZ,  the  lower  and  upper 
bases  of  the  frustum  ;  XX  ,  the  altitude  of  the  frustum. 


296  SOLID   GEOMETRY.  [Bk.  VII. 

Proposition  VI. 

402.  Theorem.  Parallel  transverse  sections  of  a  pyrami- 
dal space  are  similar  polygons,  whose  areas  are  proportional 
to  the  squares  of  the  distances  from  the  vertex  to  the  cutting 
planes. 


Given         P,  a  pyramidal   space  with  vertex   V,  cut  by  two 
parallel  planes,  B,  B',  making  transverse  sections 

ABC =  S,  A'B'C =  S',  respectively ;  VXA.  B, 

VX'  _L  B'. 

To  prove    that  (1)  S~  S', 

(2)  S:S'=  VX2 :  VX'2. 

Proof.    1.  v  the  sides  of  S  are  II  to  the  sides  of  S', 

VI,  prop.  XXIV 

.*.  VA:  VA'=  VB:  VB'  =  and  so  on  for  other 

points.  IV,  prop.  X,  cor.  2 

2.  .'.  S~  S',  which  proves  (1).       §  258,  def.  sim.  figs. 

3.  And  v  AB  :  A'B'  =  VA  :  VA'  =  VX  :  VX', 

IV,  prop.  X,  cors.  1,  3 

and  8 :  S'  =  AJi2 :  A'V><\  V,  prop.  IV 

.'.  S:  S'  =  VX*  :  VX".  IV,  prop.  VII 


Prop.  VI.]     PRISMATIC  AND  PYRAMIDAL   SPACE.  297 

Note.  The  definition  of  similar  figures,  given  in  Book  IV,  §  258,  is 
general  ;  the  center  of  similitude  and  the  given  figures  may  or  may  not 
be  in  the  same  plane.  Thus  in  the  figure  on  p.  296,  V  is  the  center  of 
similitude  of  the  triangles  ABC  and  A'B'C,  and  in  the  figure  on  p.  295, 
V  is  the  center  of  similitude  of  the  triangles  XYZ  and  ABC.  Neither 
is  the  definition  limited  to  plane  figures  ;  we  may  have  similar  solids  as 
well.  Thus  two  balls  are  similar,  or  two  cubes,  or  two  regular  tetra- 
hedra,  etc. 

Corollaries.  1.  If  a  pyramid  is  cut  by  a  plane  parallel 
to  the  base,  (1)  the  edges  and  altitude  are  divided  proportion- 
ally, (2)  the  section  is  similar  to  the  base. 

If  the  planes  in  the  proof  on  p.  296  are  on  the  same  side  of  F,  step  3 
proves  (1),  and  step  2  proves  (2).     Or,  in  the  figure  on  p.  295, 

VV'-.XX'  =  VA  :XA, 

and  A  ABC  ^  A  XYZ. 

2.  In  pyramids  having  equal  bases  and  equal  altitudes, 
transverse  sections  parallel  to  the  bases,  and  equidistant  from 
them,  are  equal ;  if  the  bases  are  congruent,  so  are  the  sections. 

1.  Let  s,  s'  be  the  areas  of  the  sections,  b,  b'  the  areas  of  the  bases,  d 
the  distance  of  the  section  from  the  vertex,  and  a  the  altitude. 

2.  Then  from  prop.  VI, 

s  :  b  =  d2  :  a2, 

and  s' :  b'  =  d2  :  a2. 

.-.  s:b  =  s' :  b'.  Ax.  1 

3.  But  b  =  b\ 

.-.  s  =  s'. 

4.  And  if  the  bases  are  congruent,  so  are  the  sections,  since  they  are 
both  similar  and  equal  to  the  bases. 

3.   The  bases  of  a  frustum  of  a  pyramid  are  similar  figures. 

For  they  are  parallel  transverse  sections  of  a  pyramidal  space.  Hence 
step  2,  p.  296,  proves  the  corollary. 


Exercise.  635.  In  the  figure  on  p.  296,  suppose  Z.  ABC  a  right  angle, 
AB  =  3  in.,  AC  =  5  in.,  VB  =  10  in.,  and  the  area  of  A  A'B'C'  =  12  sq.  in. ; 
find  the  length  of  VB'. 


298  SOLID   GEOMETRY.  [Bk.  VII. 


4.     THE  MENSURATION   OF   THE  PRISM. 

403.    Definition.     The  area  of  the  prismatic    surface  of  a 
prism  is  called  the  lateral  area  of  the  prism. 

Similarly  for  the  pyramid,  and  for  the  cylinder  and  cone,  to  be  defined 
hereafter. 


Proposition  VII. 


404.    Theorem.     The    lateral  area  of  a  prism  equals  the 
product  of  an  edge  and  the  perimeter  of  a  right  section. 


Given         the  prism  P;  a  right  section  R  with  sides  su  s2, ; 

f,f2, the  faces  of  the  prism;  e,  an  edge. 

To  prove    that  the  lateral  area  of  P  is  e  •  (st  +  s2  + ). 

Proof.    1.  '  • '  by  definition  of  right  section,  B  _L  e, . ' .  sx  _L  e.  §  339 

2.  V/u  f*> are  UJ,  Prop.  V,  cor.  2 

.'.  area/!  =  e-sx.  V,  prop.  II,  cor.  3 

3.  And  V  the  edges  are  equal,  Prop.  V,  cor.  3 
.'.  area  f2  —  e-s2,  and  so  for  the  other  faces. 

4.  .*.  lateral  area=  e\«i  +  e-  s2  + ==  e  •  (st  +  s2  + )• 

Ax.  2 


Sec.  405.]       THE  MENSURATION    OF   THE  PRISM.  299 

405.  Definitions.  A  prism  whose  edges  are  perpendicular 
to  the  base  is  called  a  right  prism  ;  if  the  edges  are  oblique  to 
the  base  it  is  called  an  oblique  prism. 

E.g.  on  p.  301  R  is  a  right  prism  and  0  is  an  oblique  prism.  So  a 
cube  is  a  special  kind  of  a  right  prism,  and  the  parallelepiped  illustrated 
on  p.  290  is  an  oblique  prism. 

The  distance  from  any  point  in  one  base  of  a  prism  to  the 
plane  of  the  other  base  is  called  the  altitude  of  the  prism. 
Similarly  for  a  parallelepiped,  which  is  a  special  kind  of  prism. 

Corollary.     The  lateral  area  of  a  right  prism  equals  the 
product  of  the  altitude  and  the  perimeter  of  the  base. 
For  the  altitude  here  equals  the  edge. 


Exercises.  636.  Required  the  lateral  area  of  a  prism  of  edge  3  in., 
the  right  section  being  an  equilateral  triangle  of  area  i  V3  sq.  in.  Also 
the  lateral  area  of  one  of  edge  3  in. ,  the  right  section  being  a  square  of 
diagonal  V2  in. 

637.  Required  the  lateral  area  of  a  right  prism  whose  base  is  a  square 
of  area  9  sq.  in.,  and  whose  altitude  equals  the  diagonal  of  the  base. 
Also  required  the  total  area. 

638.  Required  the  total  area  of  a  right  prism  whose  base  is  an  equi- 
lateral triangle  of  area  ±  V3,  and  whose  altitude  equals  a  base  edge. 

639.  Required  the  total  area  of  a  right  prism  whose  base  is  a  regular 
hexagon  whose  side  is  1  in.,  the  altitude  of  the  prism  being  equal  to  the 
diameter  of  the  circle  circumscribing  the  base. 

640.  Required  the  lateral  area  of  a  prism  of  edge  |,  the  right  section 
being  a  regular  hexagon  of  area  §  Vs. 

641.  Required  the  total  area  of  the  prism  mentioned  in  ex.  640, 
supposing  it  to  be  a  right  prism. 

642.  A  converse  of  prop.  VI  is  as  follows :  If  two  similar  polygons 
have  their  corresponding  sides  parallel,  and  lie  in  different  planes,  the 
lines  through  their  corresponding  vertices  are  concurrent.  Prove  it. 
(A  generalization  of  the  idea  of  similar  figures  in  perspective  ;  see  the 
definition  of  similar  figures  §  258,  and  the  note  at  the  top  of  p.  297.) 

643.  Investigate  and  prove  whether  or  not  any  three  faces  of  a  tetra- 
hedron are  together  greater  than  the  fourth. 


300  SOLID   GEOMETRY.  [Bk.  VII. 

Proposition  VIII. 

406.    Theorem.     Prisms  cut  from  the  same  prismatic  space 
and  having  equal  edges  are  equal. 


Given  two  prisms,  P,  P',  cut  from  the  same  prismatic 
space  S,  and  having  equal  edges  e. 

To  prove    that  P  =  P. 

Proof.  1.  If  7T=the  portion  of  the  prismatic  space  between 
P  and  P,  then  by  adding  e  to  the  edges  of  K,  each 
edge  of  P  +  K=  an  edge  of  K -f-  P.  Ax.  2 

2.  Then  '•'  P  +  K  can  evidently  slide  along  in  the 
prismatic  space  and  occupy  the  position  of  K  -f  P', 

.'.P  +  K*zK+P'.  §57 

3.  ..P  =  P.  Ax.  3 

Corollaries.  1.  Right  jyrisms  having  equal  altitudes  and 
congruent  bases  are  congruent. 

2.  An  oblique  prism  is  equal  to  a  right  prism  whose  base 
and  altitude  are  respectively  a  right  section  and  edge  of  the 
oblique  j)ris7ti. 


Prop.  IX.J        THE  MENSURATION   OF    THE   PRISM.  301 

Proposition  IX. 

407.  Theorem.  The  two  triangular  prisms  into  which  any 
parallelepiped  is  divided  by  a  plane  through  two  opposite 
edges  are  equal. 


Given  0  and  E,  parallelepipeds  with  equal  edges,  cut  from 

a  prismatic  space,  B  being  right ;  also,  P,  a  plane 
through  two  opposite  edges  of  that  space,  cutting 
R,  0  into  two  triangular  prisms,  Tx  and  T2,  Xx  and 
X2,  respectively. 

To  prove    that         (1)  Tx=  T2,   (2)  A\  =  X2. 

Proof.    1.  The  base  of  Tl  ^  the  base  of  T%.       I,  prop.  XXIV 

2.  '•'  they  have  the  same  altitude,  .".  Tx  =  T2.      Why  ? 

3.  Xi  =  -Tu  and  X2  =  T2.  Prop.  VIII 

4.  And         v  Tx  =  T2,  .'.  X,  =  X2.  Ax.  1 

Corollary.  A  trianrjular  prism  is  half  of  a  parallele- 
piped of  the  same  altitude,  whose  base  is  the  parallelogram  of 
which  one  side  of  the  triangular  base  is  the  diagonal  and  the 
other  two  are  the  sides.     (Why  ?) 


302 


SOLID   GEOMETRY. 


[Bk.  VII. 


Proposition  X. 

408.   Theorem.     Any  parallelepiped  is  equal  to  a  rectan- 
gular parallelepiped  of  equal  base  and  equal  altitude. 


Given         a  parallelepiped,  III. 

To  prove  that  III  equals  a  rectangular  parallelepiped  as  I,  of 
equal  base  and  equal  altitude. 

Proof.  1.  Let  II  be  a  parallelepiped  on  the  same  base,  />,  as 
III,  formed  by  a  rectangular  prismatic  space,  B, 
cutting  the  prismatic  space  S  of  the  figure. 

Let  I  be  a  rectangular  parallelepiped  cut  from  B, 
with  a  base  B'  =  B,  and  a  base  edge  e'  consequently 
equal  to  base  edge  e  of  II.  II,  prop.  I,  cor.  4 

2.  Then  III  =  II,  being  part  of  S  and  having  a  com- 
mon edge.  Prop.  VIII 

3.  And  I  =  II,  being  part  of  B  and  having  an  equal 
edge.  Prop.  VIII 


4. 


Ill  =  I. 


Ax.  1 


Prop.  XI.]        THE   MENSURATION   OF   THE   PRISM.  303 

Proposition  XI. 

409.    Theorem.     Two    rectangular  parallelepipeds    having 
congruent  bases  are  proportional  to  their  altitudes. 


R' 


pj^ 


Given         two  rectangular  parallelepipeds,  R  and  R',  with  alti- 
tudes a  and  a'  respectively,  and  with  bases  b. 

To  prove    that  R  :  R'  =  a  :  a'. 

Proof.    1.  Suppose  a  and  a'  divided  into  equal  segments,  I,  and 
suppose  a  =  nl,  and  a'  =  n'l. 

(In  the  figures,  n  —  6,  n'  =  4.) 
Then  if  planes  pass  through  the  points  of  division, 
parallel  to  the  bases, 

R  =  n  congruent  rectangular  ppds.  bl, 
and  R'  =  n'         "  "  "       « 

7i  -  bl      n  _a 
~ar 


2.       .-.=:  = 


hi 


Why 


Note.  The  student  should  notice  the  resemblance  between  this 
theorem  and  Bk.  V,  prop.  X.  The  above  proof  assumes  that  a  and  a' 
are  commensurable,  and  hence  that  they  can  be  divided  into  equal 
segments  I.  The  proposition  is,  however,  entirely  general.  The  proof 
on  p.  304  is  valid  if  a  and  a'  are  incommensurable. 


Exercise.     644.    Given  the  diagonals,  a,  6,  c,  of  three  unequal  faces    "»-., 
of  a  rectangular  parallelepiped,  to  compute  the  edges. 


304 


SOLID   GEOMETRY. 


[Bk.  VII. 


410.    Proof  for  incommensurable  case. 

R 


a\\\ 


1.  Suppose  a  divided  into  equal  segments  /, 

and  suppose  a  —  nl,  while  a'  =  n'l  +  some  remainder 
x,  such  that  x  <  I. 

Then  if  planes  pass  through  the  points  of  division, 
parallel  to  the  bases, 

R  =  n  congruent  rectangular  ppds.  bl, 
and  R'  =  71'         «  «  "       "    +    a  re- 

mainder bx  such  that  bx  <  bl. 


2.  Then  a'  lies  between  n'l  and  (V  +  1)  /, 
and  H'  lies  between  ?i'  •  bl  and  (n'  +  1)  bl. 

3.  .'.  —  lies  between  —  and  > 

a  n  n 

while  -=-  lies  between  —  and 

li  n  n 


Why  ? 
Why? 


a'  li'  1 

4.  .*.  —  and  —  differ  by  less  than  -■ 
a  R  J  n 


Why  ? 


5.  And  v  —  can  be  made  smaller  than  any  assumed 

difference,  by  increasing  n, 

.'.to  assume  any  difference  leads  to  an  absurdity. 


a'      R'      .  R       a 

6.  .'.  —  =  —  5  whence  —  =  — :■ 
a       R  R'      a' 


Prop.  XII.]       THE  MENSURATION   OF   THE   PRISM. 


301 


Proposition  XII. 

411.    Theorem.     Two  rectangular  parallelepipeds  of  equal 
altitudes  are  proportional  to  their  bases. 


P  0  P' 

a.  "I       a  a 


Given     »   two  rectangular  parallelepipeds,  P,  P',  having  alti- 
tudes a,  and  bases  be  and  yz,  respectively. 

To  prove    that  P  :  P'  =  bc:  yz. 

Proof.    1.   Suppose  a  rectangular  parallelepiped  Q  to  have  an 
altitude  a  and  a  base  ?/c. 

2.  Then  v  ac  may  be  considered  the  base  of  P  and  Q, 

.  ^_^ 


Q       c 
3.  And  similarly,  -^  =  — 


P 
P' 


Prop.  XI 
Why  ? 
Why  ? 


412.    Definition.     The    length,  breadth,  and  thickness   of  a 
rectangular  parallelepiped  are  called  its  three  dimensions. 


Exercise.  645.  If  through  a  point  on  a  diagonal  plane  of  a  parallele- 
piped planes  are  passed  parallel  to  the  two  pairs  of  faces  not  intersected 
by  the  diagonal,  the  parallelepipeds  on  opposite  sides  of  that  diagonal 
plane  are  equal.     (See  II,  prop.  IV.) 


306 


SOLID   GEOMETRY. 


[Bk.  VII. 


Proposition  XIII. 

m 

413.    Theorem.     Two  rectangular  parallelepipeds  are  pro- 
portional to  the  products  of  their  three  dimensions. 


Given         two   rectangular   parallelepipeds,  P,  P',   of  dimen- 
sions a,  b,  c,  and  a',  b',  c',  respectively. 

To  prove    that  P  :  P'  =  abc  :  a'b'c'. 

Proof.    1.  Suppose  a  rectangular  parallelepiped  Q  to  have  the 
three  dimensions  a',  b,  c'. 


2.  Then 

and 
3. 


Corollaries.  1.  The  volume  of  a  rectangular  parallele- 
piped equals  the  product  of  its  three  dimensions. 

This  means  that  the  number  which  represents  the  volume  is  the  product 
of  the  three  numbers  representing  the  dimensions.  That  is,  the  number 
of  times  the  unit  of  volume  is  contained  in  the  given  parallelepiped,  is  the 
product  of  the  numbers  of  times  the  unit  of  length  is  contained  in  three 
concurrent  edges. 

If  T"  is  a  cube,  of  edges  1,  1,  1;  then  P'  is  the  unit  of  measure  of 
volume.  But  P  :  P'  then  becomes  7*  :  1 ,  and  abc  s  1  •  1  •  1  theu  becomes 
abc  :  1.     .«.  P:  1  =  abc  :  1,  or  P  =  abc. 


P       ac 
Q~  a'cr 

Prop.  XII 

Q  _b 
P'~  b'' 

Prop.  XI 

P        abc 

P'  ~  a'b'c' 

Why  ? 

Prop.  XIII.]      THE   MENSURATION    OE    THE   PRISM.  307 

2.  The  volume  of  any  parallelepiped  equals  the  product  of 
its  base  and  altitude. 

For  (prop.  X)  it  equals  a  rectangular  parallelepiped  of  equal  base  and 
equal  altitude. 

3.  The  volume  of  a,  triangular  prism  equals  the  product  of 
its  base  and  altitude. 

Cor.  2  with  prop.  IX,  cor.     Let  the  student  give  the  proof  in  detail. 

4.  The  volume  of  any  prism  equals  the  product  of  its  base 
and  altitude. 

For  it  can  be  cut  into  triangular  prisms  by  diagonal  planes  through 
a  lateral  edge,  the  sum  of  the  bases  of  the  triangular  prisms  being  the 
base  of  the  given  prism.  .-.  cor.  3  applies.  Let  the  student  draw  the 
figure  and  give  the  proof  in  detail. 

5.  Any  prism  equals  a  rectangular  parallelepiped  of  equal 
base  and  equal  altitude. 

Cors.  4,  2. 

6.  The  volume  of  an  oblique  prism  equals  the  product  of  an 
edge  and  a,  right  section. 

Cor.  4  with  prop.  VIII,  cor.  2. 

7.  Prisms  having  equal  bases  are  proportional  to  their  alti- 
tudes. 

For  if  a  is  the  altitude  and  b  the  base,  then  P  =  ab,  and  P'  =  a'b'. 
If  b  —  &',  then  P'  =  a'b.      Hence    P  :  P'  —  ab  :  a'b  —  a-.a'. 

8.  Prisms  having  equal  altitudes  are  proportional  to  their 
bases.     Prisms  having    equal   bases  and  equal  altitudes   are 

equal. 

Let  the  student  give  the  proof. 


Exercises.     646.    What  is  the  edge  of  the  cube  whose  volume  equals 
that  of  a  rectangular  parallelepiped  with  edges  2.4  m,  0.9  m,  0.8  m  ? 

647.  From  the  given  edge  e  of  a  cube,  compute  (1)  the  cube's  entire 
surface,  (2)  its  diagonal,  (3)  its  volume. 

648.  Draw  a  figure  illustrating  geometrically  the  formula 

(a  +  bf  =  a3  +  &3  +  3  a2b  +  3  ab2. 


308 


SOLID   GEOMETRY 


[Bk.  VII. 


5.     MENSURATION  OF  THE  PYRAMID. 

414.  Definitions.  A  regular  pyramid  is  a  pyramid  whose 
base  is  a  regular  convex  polygon,  the  perpendicular  to  which, 
at  its  center,  passes  through   the  vertex  of 

the  pyramid. 

415.  The  slant  height  of  a  regular  pyramid 
is  the  distance  from  the  vertex  to  any  side  of 
the  base. 

E.g.  VB  in  the  annexed  figure. 

416.  The  portion  of  the  slant  height  of  a 
regular  pyramid  cut  off  by  the  bases  of  a 
frustum  is  called  the  slant  height  of  the 
frustum. 

Corollary.  The  slant  height  of  a  regular  pyramid,  or  of 
a  frustum  of  a  regular  pyramid,  is  the  same  on  ivhat ever  face 
it  is  measured. 

Let  the  student  show  that  the  faces  are  all  congruent ;  hence  that  the 
slant  heights  are  equal. 

Exercises.  649.  To  pass  a  plane  through  a  given  pyramid  parallel  to 
the  base,  so  that  the  section  shall  equal  half  the  base. 

650.  The  edges  of  a  rectangular  parallelepiped  are  3,  4,  5 ;  required 
the  total  area  of  the  faces,  the  areas  of  its  diagonal  planes,  the  length  of 
its  diagonal  line,  andjthe  lengths  of  the  diagonals  of  its  faces.  Similarly 
for  a  cube  of  edge  V2. 

651.  If  a  cubic  block  of  sandstone  at  a  temperature  of  0°  Centigrade 
has  an  edge  1  m  long,  and  if  for  every  1°  Centigrade  increase  of  tempera- 
ture the  edge  increases  0.000012  of  its  length  at  0°,  find  the  volume  at  40° 
Centigrade. 

652.  A  brick  has  the  dimensions  25  cm,  12  cm,  6  cm,  but  on  account 
of  sbrinkage  in  baking,  the  mold  is  27.5  cm  long,  and  proportionally 
wide  and  deep.  What  per  cent  does  the  volume  of  the  brick  decrease  in 
baking  ? 


Prop.  XIV.]        MENSURATION   OF   TUE   PYRAMID.  309 

Proposition  XIV. 

417.  Theorem.  The  lateral  area  of  the  frustum  of  a 
regular  pyramid  equals  half  the  product  of  its  slant  height 
and  the  sum  of  the  perimeters  of  its  bases. 


Given  BB',  a  frustum  of  a  regular  pyramid,  h  its  slant 
height,  s  a  side  of  base  B  and  p  its  perimeter,  s'  a 
side  of  base  B'  and  p'  its  perimeter,  I  the  lateral 
area. 

To  prove    that  I  =  \  h  (jp  +  p*). 

Proof.    1.  The  area  of  each  face  =  \h  (s  +  $')- 

V,  prop.  IT,  cor.  5 
2.  Adding  all  the  faces,   and  remembering  that  p  is 
the  sum  of  the  sides  s,  and  p'  of  the  sides  s',  we 
have  I  =  ih(p  +pr). 

Corollary.     The  lateral  area  of  a  regular  pyramid  equals 

half  the  product  of  its  slant  height  and  the  perimeter  of  its  base. 

For  in  the  above  theorem,  let  B'  =  0 ;  then  s'  and  p'  =  0  ;  .-.  I  =  }  hp. 


Exercises.  653.  Prove  the  above  corollary  independently  of  the 
theorem. 

654.  What  is  the  lateral  area  of  a  regular  pyramid  whose  base  is  a 

a 
triangle  of  altitude  %  V 3,  and  whose  slant  height  is  a  ? 

655.  What  is  the  total  area  of  a  frustum  of  a  regular  hexagonal 
pyramid  whose  base  edges  are  respectively  3  —  v3  and  3  +  v3,  and 
whose  slant  height  is  10  ? 


310 


SOLID   GEOMETRY. 


[Bk.  VII. 


Proposition  XV. 

418.    Theorem.     Pyramids  having  equal   bases  and  equal 
altitudes  are  equal. 

V 


Given 


pyramids    VAZ,    V'A'Z',    having   equal    bases,  and 
having  equal  altitudes  h. 


To  prove    that  pyramid  VAZ  =  pyramid  YA'ZK 

Proof.    1.  Suppose  their  bases  in  the  same  plane  M,  and  their 
vertices  on  the  same  side  of  M. 

Suppose  their  altitude  It,  divided  into  n  equal  parts 
and  planes  passed  through  the  division-points  par- 
allel to  31. 

Then  these  planes  will  make  equal  corresponding 
transverse  sections  because  the  bases  are  equal. 

Prop.  VI,  cor.  2 

2.  Suppose  planes  passed  through  the  sides  of  these 
sections  parallel  to  an  edge  of  the  pyramid,  making 

a   set   of    prisms  in    each  pyramid,   A,  B, and 

A',  B>, 

3.  Then  V  A  =  A', 

and  B  =  B', Prop.  XIII,  cor.  8 

.'.  A  +  B  + =  A'  +  B'+ Ax.  2 


Prop.  XV.]  MENSURATION   OF    THE  PYRAMID.  311 

4.  But  if  n  increases  indefinitely, 

A  +  B  + =pyr.   VAZ, 

and      A'  +  B'  +  =  pyr.   VA'Z*. 

5.  /.  pyr.  VAZ  =  pyr.  VAZ'.    IV,  prop.  IX,  cor.  1 

Corollaries.  1.  A  pyramid  having  a  parallelogram  for 
its  base  is  divided  into  equal  pyramids  by  a  plane  through  its 
vertex  and  two  opposite  vertices  of  the  base. 

For  the  two  pyramids  have  equal  bases  and  a  common  altitude. 

2.  A  pyramid  having  a  parallelogram  for  its  base  equals 

twice  a  triangular  pyramid  of  the  same  altitude,  tvhose  base 
equals  half  that  parallelogram.      (Why?) 

3.  A  triangular  pyramid  can  be  constructed  equal  to  any 
given  n -gonad  pyramid. 

II,  prop.  XII,  and  this  theorem. 


Exercises.  656.  Find  the  area  of  the  entire  surface  of  a  regular 
tetrahedron  of  altitude  h. 

657.  Find  the  altitude  of  a  regular  tetrahedron  of  total  area  a. 

658.  Find,  by  §  417,  the  total  area  of  a  cube  of  edge  e. 

659.  What  is  the  length  of  the  base  edge  of  a  regular  triangular 
pyramid  which  is  equal  to  a  regular  hexagonal  pyramid  of  the  same 
altitude,  the  base  edge  being  1  ? 

660.  In  prop.  XIV,  cor.,  B'  was  supposed  to  decrease  to  0  ;  supposing^ 
instead,  that  B'  increases  until  it  equals  B,  show  that  step  2  of  the 
theorem  gives  the  usual  formula  for  the  lateral  area  of  a  prism. 

f        661.    Prove  that  frustums  of  pyramids  having  equal  bases  and  equal 
altitudes,  which  themselves  have  equal  altitudes,  are  equal. 

662.  A  pyramid  has  for  its  base  a  regular  hexagon  with  its  shorter 
diagonal  Vs ;  the  altitude  equals  the  longer  diagonal;  required  the 
lateral  area  of  the  pyramid. 

663.  Find  the  total  area  of  the  pyramid  mentioned  in  ex.  662. 

664.  The  lower  base  of  a  frustum  of  a  regular  pyramid  is  a  square 
of  area  s2  ;  the  area  of  the  upper  base  is  half  that  of  the  lower  one  ;  the 
slant  height  is  s  ;  required  the  lateral  area. 


312  SOLID   GEOMETRY.  [Bk.  VII. 

Proposition  XVI. 

419.   Theorem.     A  triangular  prism  can   be   divided  into 
three  equal  triangular  pyramids. 


D 

F 

E 

D 

D 

F 

^s^/1 

^ 

^ 

1 

k< 

4g 

>^ 

NA 

/    A^ 

/C  A 

C 

A 

Given         ABCDEF,  a  triangular  prism. 

To  prove  that  ABCDEF  can  be  divided  into  three  equal  tri- 
angular pyramids. 

Proof.  1.  v  A,  E,  C  determine  a  plane,  also  C,  D,  E,  §  330,  1 
.*.  ABCDEF  =  three  triangular  pyramids,  viz., 
E-ABC,  E-ACD,   C-DEF.  Ax.  8 

2.  But  v  AABC^ADEF, 

. ' .  E-AB  C  =  C-DEF.  Prop.  XV 

3.  And  C-DEF  =  E-DCF  =  E-ACD,  because  they  have 
a  common  altitude  from  E  to  plane  ACFD,  and 
equal  bases.  Prop.  XV 

4.  . ' .  E-AB  C  =  C-DEF  =  E-A  CD.  Ax.  1 

Corollaries.     1.  A  triangular  pyramid  is  one-third  of  a 

triangular  prism  of  the  same  base  and  same,  altitude. 
For  the  prism  is  three  times  the  pyramid. 

2.  Any  pyramid  is  one-third  of  a  prism  of  the  same  base  and 
same  altitude. 

For,  dividing  the  base  into  triangles  by  drawing  diagonals,  the  pyra- 
mid may  be  considered  as  made  up  of  triangular  pyramids,  each  of 
which  is  a  third  of  a  triangular  prism  of  the  same  base  and  same  alti- 
tude ;  hence  the  sum  of  the  triangular  pyramids,  or  the  given  pyramid, 
equals  one-third  the  sum  of  the  triangular  prisms,  or  one-third  of  a 
prism  of  the  same  base  and  same  altitude. 


Sec.  420.]  MENSURATION   OF   THE   PYRAMID.  313 

3.  The  volume  of  a  pyramid  equals  one-third  the  product  of 
its  base  and  altitude. 

Cor.  2,  and  prop.  XIII,  cor.  4. 

4.  Pyramids  having  equal  bases  are  proportional  to  their 
altitudes;   having  equal  altitudes,  to  their  l>ases. 

-^  ,    t         t     /      ,    ,w     i       V        %ab        ab 

For  if  p  =  | ao,  and  p  =|ao,  then  —  = = 

p'      %  a'b'      a'h' 

,  ■<■  ,      t/     ,  ab       a 

And  it  6  =  o  r  then  =  —  • 

a'6'      a' 

Or  if  a  =  a  ,  then  =  — . 

a'b'      b' 

Or  if  a.  —  a'  and  b  =  b\  then  —  =  1,  orp  =  p',  as  stated  in  prop.  XV. 
P' 

420.  Definitions.  A  polyhedron  which  has  for  bases  any 
two  polygons  in  parallel  planes,  and  for  lateral  faces  triangles 
or  trapezoids  which  have  one  side  in  common  with  one  base 
and  the  opposite  vertex  or  side  in  common  with  the  other 
base,  is  called  a  prismatoid. 

The  altitude  of  a  prismatoid  is  the  perpendicidar  distance 
between  the  planes  of  its  bases. 


Exercises.    665.  Find  the  volume  of  the  pyramid  mentioned  in  ex.  662. 

666.  A  church-tower  is  capped  by  a  regular  octagonal  pyramid  whose 
height  is  55.5  m,  and  whose  base  edge  is  4.9  m.     Required  the  volume. 

667.  A  pentagonal  pyramid  has  equal  lateral  and  base  edges,  1  in. 
Find  the  lateral  area. 

668.  Find  the  volume  of  a  cube  the  diagonal  of  whose  face  is  a  V2. 

669.  Each   face   of    a  given   triangular   pyramid    is    an   equilateral 
triangle  whose  side  is  3.     Find  the  total  area. 

670.  Find  the  volume  of  the  tetrahedron  mentioned  in  ex.  656. 

671.  Also  of  the  pyramid  mentioned  in  ex.  667. 

672.  An  edge  of  a  regular  octahedron  is  1  in.     Find  the  volume 

673.  A  pyramid  stands  on  a  square  base  of  edge  1  m  ;  the  lateral  edge 
of  the  pyramid  is  also  1  m.     Find  the  lateral  area  and  volume. 


314 


■SOLID    GEOMETRY. 


[Bk.  VII. 


Proposition  XVII. 

421.  Theorem.  The  volume  of  a  prismatoid  of  bases  b  and 
1/,  altitude  h,  and  transverse  section  m  midway  between  the 
bases,  is  expressed  by  the  formula 

v  =  -  {b  +  b'  +  4?>i). 


Fig.  1. 


Fig.  2. 


Fig.  3. 


Proof.  1.  If  any  face,  ABFE,  of  the  prismatoid  P,  is  a  trape- 
zoid, divide  it  into  two  triangles  by  a  diagonal  EB. 
Let  V  be  any  point  in  m ;  join  F  to  the  vertices  of 
P ;  then  P  will  be  divided  into  two  pyramids  (Fig.  2) 
of  bases  b,  V  and  vertex  V,  and  also  pyramids  of 
vertex  V  and  triangular  bases  ABE,  etc.  (Pig.  3.) 
Let  EB  meet  m  at  D;  call  A  TDC  mv  (Fig.  3.) 
2.  Then  the  volume  of 

r-b  =  ib-t 


and  V-b'  =  j-6'.  -■  Prop.  XVI,  cor.  3 

This  completes  Fig.  2. 

3.  Pyramid        V-ABE  =  E-CVD  +  B-C1P)  +  V-ABC. 

Ax.  8 


4.  Of  these, 
and 


Prop.  XVI,  cor.  3 


Prop.  XVII.]        MENSURATION   OF  THE  PYRAMID.  315 

5.  But  F-ABC  =  twice  V-CBD  (or  B-CVD), 

v  AABC  =  twice  A  C£D, 
having  edge        AB  =  2  •  CD,  and  a  common  altitude. 

Prop.  XV I.  cor.  4 

6.  .-.  V-ABC  =  |  /«!•-/ 

7 .  .  • .  py  ram  id    V- ABE  =  -  •  4  %  A  x  s .  2,  8 
and  .".  the  sum  of  the  pyramids  of  the  form  of 

V-ABE  =  ^±m.  Axs.  23  8 

8.  .-.P  =  |(ft'  +  ft'  +  4i»).       Axs.  2.  8 

u 

Note.     The  Prismatoid  Formula,  o  =  -  (6  +  //  +  4  ???),  is  of  great  value 

G 
in  the  mensuration  of  solids.     From  it  can  be  derived  formulae  for  the 
volumes  of  all  of  the  solids  of  elementary  geometry.  f^.- 

Coeollary.     The  volume  of  the  frustum  of  a  pyramid,  of 

bases  b,  b'.  and  altitude  h,  is  -(b  +  b'  4-  Vbb'). 

o 

Fur  if  e,  e'  are  corresponding  sides  of  6,  b\  then  |  (e  -f  e')  is  the  corre- 
sponding side  of  to.     (Why  ?) 

e              %/&          ,        e'  ^ 

•'■  -n -  =  —7=  i  and  — 7-  =  -7=  •     0  ,  prop.  IV.) 

,  and  .-.  2  Vwi  =  V  6  +  v&'. 


i  (e  +  e')  Vm 

.-.  4to  =  6  +  6'  +  2  VW,  which  may  be  substituted  in  the  Prismatoid 
Formula. 


Exercises.  674.  By  letting  (1)  b'  =  0,  and  (2)  b'  —  b,  show  that 
(1)  prop.  XVI,  cor.  3,  and  (2)  prop.  XIII,  cor.  4,  follow,  as  special 
cases,  from  the  Prismatoid  Formula. 

675.  Calling  a  prismatoid  whose  lower  base  b  is  a  rectangle  of  length  I 
and  width  t*>,  and  whose  upper  base  b'  is  a  line  e  parallel  to  a  base  edge, 
and  whose  altitude  is  h,  a  icedge,  find  a  formula  for  the  volume  of  a 
wedge. 


316  SOLID   GEOMETRY.  [Bk.  VII. 

EXERCISES. 

676.  The  base  of  a  wedge  is  4  by  6,  the  altitude  is  5,  and  the  edge  e 
is  3.     Find  the  volume.     (See  ex.  675  )     Also,  when  e  =  0. 

677.  The  altitude  of  a  pyramid  is  divided  into  five  equal  parts  by 
planes  parallel  to  the  base.  Find  the  ratios  of  the  various  frustums  to 
one  another  and  to  the  whole  pyramid. 

678.  Two  pyramids,  P,  P',  have  square  bases,  and  are  such  that  the 
altitude  of  P  equals  twice  the  altitude  of  P',  but  the  base  edge  of  P  is 
half  as  long  as  the  base  edge  of  P'.     Find  the  ratio  of  their  volumes. 

679.  Find  the  volume  of  a  cube  whose  diagonal  is  Vo. 

680.  A  frustum  of  a  pyramid  has  for  its  bases  squares  whose  sides  are 
respectively  0.6  m,  0.5  m;  the  altitude  of  the  frustum  is  0.9  m.  Find 
the  volume. 

681.  Given  the  volume  v,  and  the  bases  6,  &',  of  a  frustum  of  a 
pyramid,  to  find  a  formula  for  (1)  its  altitude,  (2)  the  altitude  of  the 
whole  pyramid. 

682.  A  granite  monument  is  in  the  form  of  a  frustum  of  a  square 
pyramid,  surmounted  by  a  pyramid  ;  the  sides  of  the  bases  of  the  frustum 
are  1  m  and  0.8  m,  and  the  altitude  of  the  frustum  is  1.8  m ;  the  altitude 
of  the  pyramidal  top  is  0.45  m.  A  cubic  meter  of  water  weighs  a  metric 
ton,  and  granite  is  three  times  as  heavy  as  water.  Find  the  weight  of  the 
monument. 

683.  An  excavation  1.5  m  deep,  rectangular  at  top  and  bottom,  and 
in  the  form  of  a  frustum  of  a  pyramid,  has  its  upper  base  10  m  wide  and 
16  m  long,  and  the  lower  base  7.5  m  wide.  How  many  cubic  meters  of 
earth  would  it  take  to  fill  it  to  a  depth  of  0.75  m  ? 

684.  The  volume  of  a  cube  is  six  times  that  of  the  regular  octahedron 
formed  by  joining  the  centers  of  the  faces  of  the  cube. 

685.  Find  the  volume  of  a  prismatoid  of  altitude  3.5  cm,  the  bases 
being  rectangles  whose  corresponding  dimensions  are  3  cm  by  2  cm,  and 
3.5  c^n  by  5  cm. 

686.  It  is  usual  to  find  the  volume  of  a  pile  of  broken  stones  by  taking 
the  product  of  the  altitude  and  the  area  of  a  transverse  mid-section. 
Compare  this  with  the  Prismatoid  Formula  and  find  what  relation  it 
assumes  between  m  and  b  -f  b'.  Is  this  relation  true  in  the  case  of  a 
pyramid  ? 

687.  The  volume  of  a  pyramid  equals  the  product  of  the  altitude  and 
a  transverse  section  (parallel  to  the  base)  how  far  from  the  vertex  ? 


BOOK    VIII. —  THE   CYLINDER,    CONE,    AND 
SPHERE.      SIMILAR    SOLIDS. 


1.     THE   CYLINDER. 


422.  Definitions.  A  curved  surface  is  a  surface  no  part  of 
which  is  plane. 

The  number  of  kinds  of  curved  surfaces  is  unlimited,  just  as  the 
number  of  kinds  of  curves  in  a  plane  is  unlimited.  But  as  among  plane 
curves  the  circumference  is  the  best  known,  so  there  are  certain  curved 
surfaces  which  are  better  known  than  others,  and  these  are  treated  in 
this  book. 

423.  A  cylindrical  surface  is  a  surface  generated  by  a  straight 
line,  called  the  generatrix,  which  moves  so  as  constantly  to 
pass  through  a  given  curve,  called  the  directrix,  and  to  remain 
parallel  to  its  original  position. 

The  surface  of  a  piece  of  straight  pipe,  or  the  surface  of  the  paper 
in  a  roll,  is  an  example. 


424.  A  straight  line  in  any  position  of 
the  generatrix  is  called  an  element  of  the 
cylindrical  surface. 

425.  If  the  directrix  is  a  closed  curve, 
the  cylindrical  surface  incloses  a  space  of 
unlimited  length,  called  a  cylindrical  space. 

426.  A  section  of  a  cylindrical  space, 
made  by  a  plane  cutting  its  elements,  is 
called  a  transverse  section.  If  it  is  perpen- 
dicular to  the  elements,  it  is  called  a  right 
section. 

317 


One  form  of  a  cylindric- 
al surface.  AB  CBD, 
the  directrix  ;  BB',  an 
element  ;  BCB',a.  por- 
tion of  a  cylindrical 
space. 


318  SOLID   GEOMETRY.  [Bk.  VIII. 

As  a  transverse  section  of  a  prismatic  space  may  be  a  convex,  con- 
cave, or  cross  polygon,  so  a  transverse  section  of  a  cylindrical  space 
may  be  a  curve  of  any  shape  if  only  its  end-points  meet.  All  theorems, 
if  the  signs  are  properly  considered,  will  be  seen  to  apply  to  each  of  the 
three  forms  of  transverse  section,  corresponding  to  convex,  concave,  and 
cross  polygons.  The  third  is,  however,  too  complex  for  treatment  in 
elementary  works. 

427.  The  portion  of  a  cylindrical  space  included  between 
two  parallel  transverse  sections  is  called  a  cylinder. 

E.g.  the  portion  between  planes  P  and  P'  in  the  figure  on  p.  310. 
•  The  terms  bases  and  altitude  of  a  cylinder  will  be  understood,  without 
further  definition,  from  the  corresponding  definitions  under  the  prism. 
The  student  should,  throughout  this  section,  notice  the  relation  of  cylin- 
drical spaces  to  prismatic  spaces. 

A  cylinder  is  considered  as  having  the  same  directrix  as  its 
cylindrical  space,  and  as  having  for  elements  the  segments  of 
the  elements  of  the  cylindrical  surface  included  between  its 
bases. 

A  cylinder  is  said  to  be  right  or  oblique  according  as  its 
elements  are  perpendicular  or  oblique  to  the  bases. 

If  the  base  of  a  cylinder  is  a  circle,  the  cylinder  is  said  to 
be  circular. 

428.  Postulate  of  the  Cylinder.  A  cylindrical  surface  may 
be  constructed  with  any  directrix  and  with  any  original  posi- 
tion of  the  generatrix. 

In  solid  geometry  constructions  are  allowed  which  require  other 
instruments  than  the  compasses  and  straight-edge.  For  example,  this 
postulate  requires  the  generatrix  to  move  constantly  parallel  to  its  origi- 
nal position,  a  construction  manifestly  impossible  by  the  use  of  merely 
these  two  instruments. 


Exercises.    688.    Draw  a   figure   of  a   convex  cylinder;   a  concave 

cylinder  ;  a  cross  cylinder. 

689.    Prove  that  if  a  transverse  section  of  a  cylindrical  space  is  per- 
pendicular to  one  element  it  is  a  right  section. 


Prop.  I.]  THE    CYLINDER.  319 

Proposition  I. 

429.    Theorem.     Parallel    transverse   sections    of  a   cylin- 
drical space  are  congruent. 


Given  a  cylindrical  space  S,  cut  by  two  parallel  planes, 
P,  P',  so  as  to  form  two  transverse  sections,  L,  L'. 

To  prove    that  L  =  L'. 

Proof.  1.  Let  AA',  BB\  CC  be  segments  of  elements  between 
P  and  P\  0  any  point  in  P,  and  00'  II  A  A',  meeting 
P'  at  0' ;  join  0  to  A,  B,  C,  and  0'  to  A',  B',  C". 

2.  Then  00',  A  A'  determine  a  plane.   VI,  prop.  I,  cor.  2 

3.  And  v  OA  I!  O'A',   OB  II  0'5',   OC  II  O'C,  §  367 
.-.Z^O#  =  Z^'0'.£'3  ZAOC  =  ZA'0'C, §337 

4.  Also,  OA  =  0'^',   0^  =  O'B', T,  prop.  XXIV 

.'.if  L  is  placed  on  V  so  that  0  falls  on  0'  and  (L4 
lies  on  O'A',  A  will  fall  on  A',  B  on  £',  etc. 

5.  Similarly,  for  every   point  of  L   there  is   a  single 
corresponding  point  of  L'  on  which  it  will  fall. 

.'.  the  figures  are  congruent.       §  57,  def.  congruence 

Corollaries.     1.   The  bases  of  a  cylinder  are  congruent. 
•    2.    The  elements  of  a  cylinder  are  equal.      (Why  ?) 


320 


SOLID   GEOMETRY. 


[Bk.  VIII. 


Proposition  II. 

430.    Theorem.      Cylinders  cut  from  the  same  cylindrical 
space,  and  having  equal  elements,  are  equal. 


Given         two  cylinders,  AD,  A'D',  cut  from  the  same  cylin- 
drical space  S,  and  having  equal  elements  AC,  A'C 

To  prove    that  AD  =  A'D'. 

Proof.    1.  ■.'AC  =  A'C'J 

and  A'C  =  A'C, 

.:AA'=CC'.  Ax.  3 

2.  Similarly  for  BB'  and  DD',  and  for  all  other  seg- 
ments of  the  same  elements,  included  between  AB, 
A'B',  and  CD,  CD'. 

3.  And  v  CD^AB,  and  C'D'^A'B',  Prop.  I 
.*.  solid  CD'  can  be  made  to  slide  along  in  S  and 
coincide  with  solid  AB',  since  they  are  equal  in  all 
their  parts. 

4."  Adding  the  common  part  A'D, 

AD  =  A'D'.  Ax.  2 

Corollary.  The  cylindrical  surfaces  of  two  cylinders  cut 
from  the  same  cylindrical  space,  and  having  equal  elements, 
are  equal. 

For  it  is  proved,  in  step  3,  that  they  can  be  made  to  coincide. 


Secs.  431-434.] 


THE   CONE. 


321 


2.     THE   CONE. 

431.    Definitions.     A  conical  surface  is  a  surface  generated  by 
a  straight  line  which  moves  so  as  constantly  to  pass  through 
a   given   curve    and    contain    a    given    point 
called  the  vertex. 

The  terms  generatrix,  directrix,  elements  will  be 
understood  from  §§  423,  424. 

432.  The  portions  of  the  conical  surface 
on  opposite  sides  of  the  vertex  are  called  the 
nappes,  and  are  usually  distinguished  as  upper 
and  lower. 


433.  If  the  directrix  is  a  closed  curve,  the 
conical  surface  incloses  a  double  space,  on 
opposite  sides  of  the  vertex,  known  as  a 
conical  space. 

A  section  of  a  conical  space  made  by  a 
plane  cutting  all  of  its  elements  on  the  same 
side  of  the  vertex  is  called  a  transverse  section. 


A  conical  surface. 
DX,  the  directrix; 
V,  the  vertex  ;  N. 


X',  the  lower  and 
upper  nappes; 
V-JJX,  a  cone,  with 
base  the  closed 
figure  DX. 


434.  The  portion  of  a  conical  space  included  between  the 
vertex  and  a  transverse  section  is  called  a  cone,  the  transverse 
section  being  called  its  base. 

A  cone  is  considered  as  having  the  same  directrix  and  vertex 
as  its  conical  space,  and  the  segments  of  the  elements  between 
the  vertex  and  base  are  called  the  elements  of  the  cone. 

The  distance  from  the  vertex  of  a  cone  to  the  plane  of  the 
base  is  called  the  altitude  of  the  cone. 

If  the  base  of  a  cone  is  a  circle,  the  cone  is  said  to  be 
circular.  In  that  case  the  line  determined  by  the  vertex  and 
the  center  of  the  base  is  called  the  axis  of  the  cone.  If  this 
axis  is  perpendicular  to  the  base,  the  cone  is  called  a  right 
circular  cone ;  if  oblique,  an  oblique  circular  cone. 


322 


SOLID   GEOMETRY. 


[Bk.  VIII. 


A  right  circular  cone  is  often  called  a  cone  of  revolution, 
because  it  can  be  generated  by  the  revolution  of  a  right-angled 
triangle  about  one  of  its  shorter  sides.  A  right  circular 
cylinder  is  often  called  a  cylinder  of  revolution.     (Why  ?) 

435.  Postulate  of  the  Cone.  A  conical  surface  may  be  con- 
structed with  any  directrix  and  any  vertex. 

436.  Relation  of  Cone  and  Pyramid.     If  points  A,  B,  C, 

are  taken  on  the  perimeter  of  the  base  of  a  cone,  and  joined  to 
the  vertex  V,  and  if  planes  be  passed  through  VA  and   VB, 

VB  and  VC, ,  a  pyramid  will  be  formed,  called  an  inscribed 

pyramid. 

If  the  base  of  the  cone  is  bounded  by  a  convex  curve,  the  base  of  the 
pyramid  will  be  a  polygon  inscribed  in  it.  But  whether  the  base  is 
convex  or  not,  the  pyramid  is  called  an  inscribed  pyramid. 


Pyramids  inscribed  in  cones.    The  first  figure  is  a  right  circular  cone.    The 
inscribed  pyramids  are  indicated  by  dotted  lines.    It,  the  altitude. 

437.  If  the  base  of  the  cone  is  a  circle,  and  a  regular 
polygon  is  circumscribed  about  it,  the  planes  determined  by 
the  sides  of  the  polygon  and  the  vertex  of  the  cone  form, 
with  the  polygon,  a  pyramid  which  is  said  to  be  circumscribed 
about  the  circular  cone. 

There  are  other  forms  of  circumscribed  pyramids,  but  the  one  here 
mentioned  is  the  only  one  that  is  necessary  for  this  work. 

The  slant  height  of  a  right  circular  cone  is  denned  as  the 
slant  height  of  the  circumscribed  pyramid.     (Why  V) 


Secs.  438-440.] 


THE    CONE. 


323 


438.  If  a  pyramid  is  inscribed  in  or  circumscribed  aLoui  a 
cone,  a  transverse  section  of  the  pyramid  and  cone  cuts  off, 
toward  the  base,  a  frustum  of  a  cone  and  an  inscribed  or  circum- 
scribed frustum  of  a  pyramid. 

A 


ABC,  a  circumscribed  frustum  of  a  pyramid  ;  A'B'C,  an  inscribed  frustum  of  s, 
regular  pyramid ;  s,  the  slant  height  of  the  frustum  of  the  cone. 

The  terms  bases,  altitude,  and  lateral  surface  will  be  understood  from 
the  terms  used  with  the  pyramid  and  the  frustum  of  a  pyramid. 

439.  From  the  above  definitions  it  is  evident  that,  if  the 
inscribed  or  circumscribed  frustum  of  a  pyramid  has  equilateral 
bases,  then  if  the  number  of  lateral  faces  increases  indefinitely, 
the  frustum  of  the  pyramid,  its  bases,  and  its  lateral  surface, 
approach  as  their  respective  limits  the  frustum  of  the  cone,  its 
bases,  and  its  lateral  surface,  but  that  the  altitude  does  not 
vary.  If  a  frustum  of  a  right  pyramid  be  circumscribed  about 
the  frustum  of  a  right  circular  cone,  the  slant  height  of  the 
frustum  of  the  pyramid  may  be  called  the  slant  height  of  the 
frustum  of  the  cone.     Hence  the  following 

440.  Corollary.  If  F  is  the  frustum  of  a  cone,  and  F'  the 
inscribed  or  circumscribed  frustum  of  a  pyramid,  of  equilateral 
bases,  and  if  b1?  b2,  1,  v  are  the  bases,  lateral  surface,  and 
volume,  respectively,  of  F,  and  b/,  b2',  lf,  v'  the  bases,  lateral 
surface,  and  volume,  respectively,  of '  F',  then  if  the  number  of 
faces  of  F'  increases  indefinitely, 

V^bx,  b2'  =  l),,  l'  =  l,  v'  =  v. 


324  SOLID   GEOMETRY.  [Bk.  VIII. 

Proposition  III. 

441.  Theorem.  The  lateral  area  of  a  frustum  of  a  right 
circular  cone  equals  one-half  the  product  of  the  slant  height 
and  the  sum  of  the  circumferences  of  its  bases. 

Given  a  frustum  of  a  right  circular  cone,  I  its  lateral  area, 
cx  and  c2  the  circumferences  of  its  upper  and  lower 
bases,  respectively,  and  5  its  slant  height. 

To  prove    that  I  —  J  s(cx  +  c2). 

Proof.  1.  Let  I',  pi,  2h,  s  be  the  lateral  area,  the  perimeters 
of  the  upper  and  lower  bases,  and  the  slant  height, 
respectively,  of  the  circumscribed  frustum  F  of  a 
regular  pyramid. 

2.  Then  V  =  %s(Pl  +  p2).  VII,  prop.  XIV 

3.  But  if  the  number  of  faces  of  F  increases  indefi- 
nitely, V  =  I,  jh  ==  ('i ?  2h  ==  c2,  while  the  slant  height 
is  the  same.  §  440 

4.  .'.  I  =  £s  (cx  +  c2).         IV,  prop.  IX,  cor.  1 

Corollaries.  1.  If  the  radii  of  the  upper  and  lower  bases 
are  r1?  r2,  respectively,  then  1  =  7rs  (rL  -f-  r2). 

2.  If  r3  =  the  radius  of  the  circle  midway  between  the  bases 
of  the  frustum,  then  1  =  2  7rr3s. 

For  r3  =  (r1  +  r2)/2.     Why? 

J  3.   The  lateral  area  of  a  right  circular  cone  equals  half  the 
product  of  its  slant  height  and  the  circumference  of  the  base. 

If  the  upper  base  of  a  frustum  of  a  cone  decreases  to  zero,  what  does 
the  frustum  become  ?     At  the  same  time  what  does  C\  of  step  4  become  ? 

4.  The  lateral  area  of  a  right  circular  cylinder  equals  the 
product  of  its  altitude  and  the  circumference  of  the  base. 

If,  in  step  4,  cx  =  c2,  what  does  /  equal  ?     What  does  s  equal  ? 


Prop.  IV.]  THE    CONE.  325 

Proposition  IV. 

442.    Theorem.     The  volume  of  the  frustum  of  a  cone  of 
bases  hv  b2  and  altitude  h  is   expressed    by  the  formula 

▼  =  3  (\  +  b2  +  Vb^). 

Proof.  1.  Let  v',  h,  V>  b2  be  the  volume,  altitude,  and  bases, 
respectively,  of  an  inscribed  frustum  of  a  pyramid 
with  an  equilateral  base. 

h 


2.  Then  v'  =  -  (V  +  bj  +  VVV).       VII,  prop.  XVII 

o 

3.  But  if  the  number  of  faces  of  v'  increases  indefinitely, 
v'  =  v,  bx  —  bu  b.2'  —  b2,  while  h  is  constant.      §  440 

4.  .".  v  =  -  (bx  +  b2  +  ^s/bxb2).  IV,  prop.  IX,  cor.  1 

o 

Corollaries.  1.  If  the  frustum  is  circular,  and  the  radii 
of  b:,  b2  are  i\,  r2,  respectively,  then  v  =  \  7rh  (i\2  +  r22  +  r1r2). 

2.  If  v3  =  the  radius  of  the  circle  midway  between  the  bases 
of  a  frustum  of  a  circular  cone,  and  if  h  is  the  altitude,  and  xx, 
r2  are  the  radii  of  the  bases,  then  v  =  i  7rh  (i^2  +  r22  -f-  4  r32). 

See  prop.  Ill,  cor.  2. 

3.  The  volume  of  a  cone  of  base  b  and  altitude  h  is  expressed 
by  the  formula  v  =  \  hb. 

Let  b2  —  0  in  prop.  IV. 

4.  The  volume  of  a  circular  cone,  the  radius  of  whose  base  is 
r,  is  expressed  by  the  formula  v  =  \  7rr2h. 

5.  The  volume  of  a  cylinder  of  base  b  and  altitude  h  is 
expressed  by  the  formula  v  =  hb. 

Let  &i  =  62. 

6.  The  volume  of  a  cylinder  of  altitude  h  and  base  radius  r 
is  expressed  by  the  formula  v  =  7rr2h. 


326 


SOLID   GEOMETRY. 


[Bk.  VIII. 


3.     THE   SPHERE. 


443.    Definitions.     A  sphere  is  the  finite  portion  of  space 
bounded  by  a  surface,  which  is  called  a  spherical  surface  and  is 
such  that  all  points  upon  it  are  equidistant 
from  a  point  within  called  the  center  of  the 
sphere. 

A  straight  line  terminated  by  the  center 
and  the  spherical  surface  is  called  a  radius, 
and  a  straight  line  through  the  center, 
terminated  both  ways  by  the  spherical 
surface,  is  called  a  diameter  of  the  sphere. 

A  section  of  a  sphere  made  by  a  plane 
is  called  a  plane  section. 


A  sphere.  O, the  center. 
OA,OB,  radii.  AB,  a 
diameter. 


444.  Corollaries.  1.  A  diameter  of  a  sphere  is  equal  to 
the  sum  of  tivo  radii  of  that  sphere. 

2.  Spheres  having  the  same  radii  are  congruent,  and  con- 
versely. 

3.  A  point  is  within  a  sphere,  on  its  surface,  or  outside  the 
sphere,  according  as  the  distance  from  that  point  to  the  center 
is  less  than,  equal  to,  or  greater  than,  the  radius. 

445.  Postulates  of  the  Sphere.     (Compare  §  109.) 

1.  All  radii  of  the  same  sphere  are  equal,  and  hence  all 
ilia  meters  of  the  same  sphere  are  equal. 

2.  If  an  unlimited  straight  line  passes  through  a  point 
within  a  sphere,  it  must  cut  the  surface  at  least  twice. 

3.  If  an  unlimited  plane,  or  if  a  spherical  surface,  intersects 
a  spherical  surface,  it  must  intersect  it  in  a  closed  line. 

4.  A  sphere  has  but  one  center. 

5.  A  sphere  may  be  constructed  with  any  center,  and  with  a 
radius  equal  to  any  given  line  segment. 


Prop.  V, 


THE   SPHERE. 


327 


Proposition  V. 
446.    Theorem.     A  plane  section  of  a  sphere  is  a  circle. 


Given         a  sphere  with  center  0,  and  a  section  ABDC  made 
by  a  plane  M. 

To  prove    that  ABDC  is  a  circle. 

Proof.    1.  M  intersects  the  sphere  in  a  closed  line.        §  445,  3 

2.  Suppose  0  joined  to  two  points  AB  on  that  line, 
and  OC±3I;  draw  CA,  CB. 

3.  Then  v  A  OCB,   OCA  are  rt,  and  OC  =  OC,  and 
OB  =  OA, 

.'.  A  CBO  ^  A  CAO,  and  CB  =  CA.  §  88,  cor.  5 

So  for  any  other  points  on  the  closed  line. 

4.  .'.  ABDC  is  a  circle  and  C  is  its  center. 

§  165,  def.  O 

447.  Definitions.  A  great  circle  of  a  sphere  is  a  circle 
passing  through  its  center ;  a  small  circle,  one  not  passing 
through  its  center. 

Corollaries.  1.  The  line  determined  by  the  center  of  a 
sphere  and  the  center  of  any  small  circle  of  that  sphere  is 
perpendicular  to  that  circle. 

For  the  line  OC  from  the  center  of  the  sphere  perpendicular  to  the 
circle  has  been  proved  to  coincide  with  the  line  determined  by  the  center 
of  the  circle  and  the  center  of  the  sphere,  and  there  is  only  one  line  from 
the  center  of  the  sphere  perpendicular  to  the  circle. 


328  SOLID   GEOMETRY.  [Bk.  VIII. 

2.  Of  two  circles  of  a  sphere,  the  first  is  greater  than,  equal 
to,  or  less  than,  the  second,  according  as  its  distance  from  the 
center  is  less  than,  equal  to,  or  greater  than,  that  of  the  second. 

For  AC2  -  f2  —  OC2 ;  .-.  the  smaller  OC,  the  greater  AC,  etc. 

3.  A  great  circle  has  the  same  center  and  radius  as  the 
sphere  itself ;  hence  all  great  circles  of  a  given  sphere  are  equal. 

4.  A  great  circle  bisects  the  sphere  and  the  spherical  surface. 

For  if  the  two  parts  are  applied  one  to  the  other,  they  will  coincide  ; 
if  they  did  not,  the  definition  of  sphere  would  be  violated. 

5.  Two  great  circles  bisect  each  other. 

They  have  the  same  center,  and  hence  a  common  diameter. 

448.  The  student  should  notice  the  relation  between  the 
sphere  and  circle.     Thus  in  prop.  V  and  its  corollaries  : 

The  Circle.  The  Sphere. 

A  portion  of  a  line  cut  off  by  a  A  portion  of  a  plane  cut  off  by 

circumference  is  a  chord.  a  spherical  surface  is  a  circle. 

The  greater  a  chord,  the  less  its  The  greater  a  circle,  the  less  its 

distance  from  the  center.  distance  from  the  center. 

A  diameter  (great  chord)  bisects  A  great  circle  bisects  the  sphere 

the  circle  and  the  circumference.  and  the  spherical  surface. 

Two   diameters    (great    chords)  Two    great   circles   bisect    each 

bisect  each  other.  other. 

Hence  may  be  anticipated  a  line  of  theorems  on  the  sphere,  derived 
from  those  on  the  circle,  by  making  the  following  substitutions: 

1.    Circle,    2.    circumference,  1.  Sphere,   2.  spherical  surface, 

3.  line,    4.  chord,    5.  diameter.  3.  plane,   4.  circle,  6.  great  circle. 

449.  Definitions.  The  diameter  of  a  sphere,  perpendicular 
to  a  circle  of  that  sphere,  is  called  the  axis  of  that  circle,  and 
its  extremities  are  called  the  poles  of  that  circle. 

The  two  equal  parts  into  which  a  great  circle  divides  a 
sphere  are  called  hemispheres,  their  curved  surfaces  being 
called   hemispherical   surfaces. 

Corollary.     The  axis  of  a  circle  pusses  through  its  center. 


Prop.  VI.] 


THE   SPHERE. 


329 


Proposition  VI. 

450.  Theorem.  The  straight  lines  joining  any  tivo  points 
on  the  circumference  of  a  circle  of  a  sphere  to  one  of  the 
poles  of  that  circle  are  equal. 


Given         the  circle  ABC,  and  its  poles  P,  P';  PA,  PB  con- 
necting P  with  any  two  points  on  the  circumference. 

To  prove    that  PA  =  PB. 

Proof.    1.  v  OP  ±Q  ABC  at  C,  .'.  OP±AC  and  BC.  Why? 

2.  And  V  PC  =  PC,  and  CA  =  CB,  §  109 

'.'.  A  ACP  ^  A  BCP,  and  PA  =  PB.  Why  ? 

Corollary.  Great-circle  arcs  from  a  pole  of  a  circle  to 
points  on  the  circumference  of  that  circle  are  equal.     (Why  ?) 

451.  Definitions.  The  length  of  the  great-circle  arc  joining 
a  pole  to  any  point  on  the  circumference  of  a  circle  is  called 
the  polar  distance  of  the  circle. 

The  shorter  polar  distance  of  small  circles  is  to  be  understood. 

A  fourth  of  the  circumference  of  a  great  circle  is  called  a 
quadrant. 

Corollaries.  1.  Circles  of  the  same  sphere,  having  equal 
polar  distances,  are  equal.      (Why  ?) 

2.    The  polar  distance  of  a  great  circle  is  a  quadrant.   (Why  ?) 


330 


SOLID   GEOMETRY. 


[Bk.  VIII. 


Proposition  VII. 

452.  Theorem,  If  on  a  spherical  surface,  each  of  the 
great-circle  arcs  joining  a  point  to  two  other  points  {not  the 
extremities  of  a  diameter  of  the  sphere)  is  a  quadrant,  then 
that  point  is  a  pole  of  the  great  circle  through  those  points. 


Given  P,  A,  B,  three  points  on  a  spherical  surface,  and 
such  that  PA  =  PB  =  a  quadrant;  A,  B  are  not 
extremities  of  a  diameter  ;   0  is  the  center. 

To  prove    that  P  is  the  pole  of  the  great  circle  ABO. 

Proof.    1.  '•'  PA  =  PB  =  a  quadrant, 

.*.  Z  POA  =  Z  POB  =  a  rt.  Z.    Ill,  prop.  II,  cor.  2 

2.  .-.  PO  _L  O  ABO.      VI,  prop.  VI,  cor.  1 

3.  ,\  P  is  a  pole  of  O  ABO.       §  449,  def.  pole 


Exercises.  690.  How  many  points  on  a  spherical  surface  determine 
a  small  circle  ?     How  many,  in  general,  determine  a  great  circle  ? 

691.  Prove  that  parallel  circles  of  a  sphere  have  the  same  poles. 

692.  In  the  theorem  :  A  diameter  which  is  perpendicular  to  a  chord 
bisects  it,  make  the  substitutions  suggested  in  §  448,  and  prove  the  result- 
ing proposition. 

693.  Similarly  for  III,  prop.  VI. 

694.  What  is  the  locus  of  points  at  a  given  distance  r  from  a  fixed 
point  C  ? 


Prop.  VIII.]  THE   SPHERE.  331 

Proposition  YIII. 

453.  Theorem.  Of  all  planes  through  a  point  on  a  sphere 
the  plane  perpendicular  to  the  radius  drawn  to  that  point  is 
the  only  one  that  does  not  meet  the  sphere  again. 


Given  point  P  on  a  sphere  with  center  0,  and  M,  N,  two 
planes  respectively  perpendicular  and  oblique  to  OP 
at  P. 

To  prove  that  M  does  not  meet  the  spherical  surface  again, 
but  that  N  does. 

Proof.    1.   Let  OB  _L  X,  and  OA  be  any  oblique  to  M. 

Then  v  OP  is  oblique  to  N}  Why  ? 

.-.  OB  <  OP.  VI,  prop.  XI 

2.  And  v  OP  ±  Mt 

.-.  OA>  OP.  VI,  prop.  XI 

3.  .".  B  is  within,  and  ^1  without,  the  sphere. 

§  444,  cor.  3 

4.  .'.  N  meets  the  spherical  surface  in  more  than  one 
point.  §  445,  3 

5.  And  v  A  is  any  point  in  M3  except  P,  Step  3 
.'.  M  meets  the  surface  onlv  at  P. 


332  SOLID   GEOMETRY.  [Bk.  VIII. 

454.  Definitions.  A  plane  (or  line)  which,  meeting  a  spher- 
ical surface  in  one  point,  does  not  meet  it  again,  is  said  to 
be  tangent  to  the  sphere  at  that  point.  The  point  is  called 
the  point  of  tangency,  or  point  of  contact,  and  the  plane  (or  line) 
is  called  a  tangent  plane  (or  line). 

Corollaries.  1.  One  and  only  one  plane  can  be  passed 
through  a  given  point  on  a  sphere,  tangent  to  that  sphere. 
(Why?) 

2.  Any  tangent  plane  is  perpendicular  to  the  radius  at  the 
point  of  contact. 

For  it  cannot  be  oblique  and  be  a  tangent  plane.     Step  4. 

3.  A  plane  perpendicular  to  a  radius  at  its  extremity  on  the 
spherical  surface  is  tangent  to  the  sphere. 


k 


Exercises.  695.  To  find  a  point  in  a  given  plane,  equidistant  from 
two  fixed  points  in  that  plane,  and  at  a  given  distance  ci!  from  a  point  C 
not  in  that  plane.     Discuss  for  0,  1,  2  solutions. 

696.  Prove  that  the  lateral  area  of  any  right  cylinder  equals  the 
product  of  its  altitude  and  the  perimeter  of  the  base.  (Inscribe  a  prism 
and  apply  the  theorem  of  limits.) 

697.  How  many  square  feet  in  the  surface  of  a  cylindrical  water  tank, 
open  at  the  top,  its  height  being  40  ft.,  and  its  diameter  40  ft.  ? 

698.  Considering  the  moon  as  a  circle  of  diameter  21G0.6  miles  whose 
center  is  234,820  miles  from  the  eye,  what  is  the  volume  of  the  cone 
whose  vertex  is  the  eye  and  whose  base  is  the  full  moon  ? 

699.  Find  a  point  whose  distance  from  a  fixed  point  is  d  and  whose 
distance  from  each  of  two  intersecting  planes  is  cV.  Discuss  the  solution 
as  to  impossible  cases,  and  the  number  of  such  points  for  possible  cases. 

700.  Find  the  locus  of  points  equidistant  from  two  given  points,  and 
at  a  given  distance  d  from  a  given  point. 

701.  To  determine  a  plane  which  shall  pass 

(1)  through  a  given  line  and  be  (2)  through  a  given  point and  be 

at  a  given  distance  from  a  given       at  a  given  distance  from  a  given 
point.  line. 


Prop.  IX.] 


THE  SPHERE. 


333 


Proposition  IX. 

455.   Theorem.     Four  jjoints,   not    coplanar,   determine   a 
spherical  surface. 


Given         four  points,  A,  B,  C,  D,  not  coplanar. 

To  prove    that  A,  B,  C,  D  determine  a  spherical  surface. 

Proof.    1.  Draw  AB,  BC,  CD,  DA,  AC. 

Let  E  be  the  circumcenter  of  A  ACD,  F  of  A  ABC, 

EH  1.  ACD,  FJ±ABC. 

2.  Then  F,  F  are  on  the  _L  bisectors  of  AC;  call  these 
_L  bisectors  GF,  OF.  I,  prop.  XLI 

3.  And  v  FA  =  EC  =  ED  (Why  ?), 

.'.  any  point  on  EH  is  equidistant  from  A,  C,  D. 

VI,  prop.  XI,  3 
Similarly,  any  point  on  FJ  is  equidistant  from 
A,  B,  C. 

4.  But  CA  ±  plane  EOF.  Why  ? 

5.  .'.  planes  ABC,  ACD  _1_  EOF.  Why  ? 

G.   .'.  both  EH  and  FJ  lie  in  plane  EOF. 

VI,  prop.  XIX,  cor.  1 

7.  And  V  FJ  meets  EH,  uniquely,  as  at  P, 

I,  prop.  XVII,  cor.  4 
.*.  P  is  the  center  of  a  sphere  whose  surface  passes 
through  A,  B,  C,  D,  and  there  is  only  one  such  sphere. 


334  SOLID   GEOMETRY.  [Bk.  VIII. 

456.  Definitions.  A  sphere  is  said  to  be  circumscribed  about 
a  polyhedron  if  the  vertices  of  the  polyhedron  all  lie  on  the 
spherical  surface ;  the  polyhedron  is  then  said  to  be  inscribed 
in  the  sphere. 

Corollaries.  1.  Two  spherical  surfaces  having  four 
common  points,  not  coplanar,  coincide. 

For  by  step  7  they  have  the  same  center,  P,  and  the  same  radius. 

2.  The  perpendiculars  to  the  four  faces  of  a  tetrahedron, 
through  the  circumcenters  of  those  faces,  are  concurrent. 

For  each  of  these  perpendiculars  passes  through  P,  the  center  of  the 
sphere  whose  surface  is  determined  by  the  four  vertices. 

3.  A  sphere  can  be  circumscribed  about  any  tetrahedron. 

457.  The  angle  between  two  great-circle  arcs  is  defined  as 
being  the  plane  angle  between  tangents  to  those  arcs  at  their 
point  of  meeting. 

P 


E.g.  the  angle  made  by  arcs  AP,  BP  is  defined  as  the  plane 
angle  A'PB'. 

458.    From  this  definition  follow  these  corollaries  : 

1.  The  angle  made  by  two  arcs  has  the  same  numerical 
measure  as  the  dihedral  angle  of  their  planes.     (§  359.) 

2.  An  angle  made  by  tivo  arcs  has  the  same  numerical  meas- 
ure as  the  arc  which  these  arcs  intercept  on  the  circumference 
of  the  great  circle  of  wh  ich  the  vertex  is  the  pole. 

That  is,  Z  APB  =  /.A'PB'-  ZAOB,  which  has  the  same  numerical 
measure  as  AB. 


Secs.  459-463.] 


THE   SPHERE. 


335 


459.  A  spherical  polygon  is  a  portion  of  a  spherical  surface 
bounded  by  arcs  of  great  circles. 

The  words  sides,  angles,  vertices,  etc.,  are  used  as  with  plane  polygons. 

460.  A  spherical  polygon  is  said  to  be  convex  when  each 
side  produced  leaves  the  entire  polygon  on  the  same  hemi- 
sphere ;  otherwise  it  is  said  to  be  concave. 


In  the  figure,  ABP  is  a  convex  polygon,  for  if  any  side,  as  PB,  is 
produced  it  leaves  the  entire  polygon  on  the  hemisphere  to  the  left  of  PB. 
But  QRS  T  is  concave,  because  side  SB,  or  QR,  produced,  leaves  part  of 
the  polygon  on  one  hemisphere  thus  formed,  and  part  on  the  other. 

461.  Corollary.  No  side  of  a  convex  spherical  polygon  is 
greater  than  a  semicircumference. 

For  if  AP>  semicircumference,  suppose  XP  =  a  semicircumference. 
Then  v  great  circles  bisect  each  other  (prop.  V,  cor.  5),  PB  must  pass 
through  X ;  but  then  PB  produced  would  leave  part  of  the  polygon  on 
one  hemisphere  and  part  on  the  other,  so  that  it  could  not  be  convex. 

462.  A  lune  is  a  portion  of  a  spherical  surface  bounded  by 
the  semicircumferences  of  two  great  circles.  The  angle  of  a 
lune  is  that  angle  towrard  the  lune  made  by  the  bounding  arcs. 

In  the  figure,  PAP'B  is  a  lune,  and  Z.  APB,  or  /.  BP'A,  is  its  angle. 
The  limiting  cases  of  a  lune  are  evidently  a  semicircumference,  when 
the  angle  is  zero,  and  a  spherical  surface,  when  the  angle  is  360°. 

463.  Corollary.  Limes  on  the  same  sphere,  and  having 
the  same  angle,  are  congruent. 

For  one  can  evidently  be  made  to  coincide  with  the  other. 


336  SOLID   GEOMETRY.  [Bk.  VIII. 

Proposition  X. 

464.   Theorem.     On  the  same  sphere  or  on  equal  spheres 
lunes  are  proportional  to  their  angles. 


(In  this  figure  the  eye  is  supposed  to  be  looking  down  on  the  sphere 
from  above  the  angle  of  the  lune,  as  on  the  North  Pole  of  the  earth.  This 
allows  only  half  of  each  lune  to  be  seen.) 

Given         two  lunes,  C  and  D,  with  angles  A  and  B  respec- 
tively. 

To  prove    that  A  :  B  =  C :  D. 

Proof.    1.  If  C  and  D  are  on  different  spheres,  they  can  be 
placed  in  the  relative  positions  shown  in  the  figure. 

§  444,  cor.  2 
Suppose  A  and  B  divided  into  equal  A,  x,  and  sup- 
pose A  =  nx,  and  B  =  n'x. 

(In  the  figure  n  =  6,  n'  =  4.) 

2.  Then  C  is  divided  into  n  congruent  lunes,  y, 

and    D  "  "    n'  "  «  §  463 

3.  m*mA  =  2L=*=3L  =  l.  Why? 

B      n'x      n'      n'y      D  y' 


Exercise.  702.  The  six  planes  perpendicular  to  the  six  edges  of  a 
tetrahedron  at  the  mid-points  of  its  edges,  meet  in  a  point,  (Is  this 
point  the  center  of  a  particular  sphere  ?) 


Prop.  X.]  THE   SPHERE.  337 

465.    Proof  for  incommensurable  case.     (Compare  §  410.) 


1.  Suppose  A  divided  into  equal  A,  x,  and  suppose 
A  =  nx,  while  B  =  n'x  +  some  remainder  w,  such 
that  w  <  x. 

Then  C  is  divided  into  n  congruent  lunes,  y,  and 
D  is  the  sum  of  n'  congruent  lunes,  y,  +  a  remain- 
der z,  such  that  z  <  y. 

2.  Then  B  lies  between  n'x  and  (n*  +  1)  x,  Why  ? 
and    D    "         "         n'y    "    (%'  +  1)  */.             Why  ? 

3.  .".  —  lies  between  —  and — ,  Why? 

A  nx  nx 

...    2>v      .  ra'v       ,   (V +  1W 

while  —  lies  between  — —  and —  • 

C  ny  ny 

4.  . ' .  —  and  —  both  lie  between  —  and 

A  6  n  n 

7?  T)  1 

5.  .*.  —  and  —  differ  by  less  than  -■  Why  ? 

j\.  o  n> 

6.  And  v  -  can  be  made  smaller  than  any   assumed 

n  J 

difference,  by  increasing  n, 

.*.to  assume  any  difference  leads  to  an  absurdity. 

B      D      .  AC 

7.  /.2  =  -,  whence -==-• 


338  SOLID   GEOMETRY.  [Bk.  VIII. 

466.  Definition.  The  solid  bounded  by  a  lune  and  two 
semi-circles  is  called  a  spherical  wedge. 

The  angle  of  the  lune  is  called  the  angle  of  the  wedge. 
The  word  iingula  is  sometimes  used  for  spherical  wedge. 

Corollaries.  1.  A  lune  is  to  the  spherical  surface  on 
which  it  lies  as  the  angle  of  the  lune  is  to  a  perigon. 

For  the  spherical  surface  may  be  considered  as  a  lune  whose  angle  is 
a  perigon. 

2.  A  spherical  ivedge  is  to  the  sphere  of  which  it  is  a  part 
as  the  angle  of  the  wedge  is  to  a  perigon. 

In  the  proof  of  prop.  X,  if  we  should  substitute  the  word  wedge  for 
the  word  lune,  and  consider  the  sphere  as  a  wedge  whose  angle  is  a 
perigon,  the  corollary  would  evidently  be  proved. 


* 


Exercises.  703.  To  draw  a  plane  tangent  to  a  given  sphere,  from  a 
point  on  the  sphere.  (See  III,  prop.  XXVI.)  Also,  to  draw  one  from 
an  external  point. 

704.  To  find  the  locus  of  centers  of  spheres  whose  surfaces  (1)  pass 
through  two  given  points ;  (2)  are  tangent  to  two  given  coplanar  lines  ; 
(3)  are  tangent  to  two  given  planes.  (As  special  cases,  the  lines  may  be 
parallel  and  the  planes  may  be  parallel.) 

705.  What  is  the  locus  of  the  centers  of  spheres  whose  surfaces  (1)  pass 
through  the  vertices  of  a  given  triangle  ?  (2)  are  tangent  to  the  sides  of  a 
given  triangle  ? 

706.  To  find  the  center  of  a  sphere  whose  surface  includes  both  a 
given  circumference  and  a  point  not  in  the  plane  of  that  circumference. 
(As  a  special  case,  suppose  the  point  is  on  the  perpendicular  to  the  plane 
of  the  given  circle  through  the  center.) 

707.  In  the  figure  of  prop.  IX  show  that  E,  G,  F,  P  are  concyclic. 
Hence  show  that  six  circumferences  intersect  by  threes  in  the  ciivum- 
centers  of  the  faces  of  a  tetrahedron,  and  all  intersect  in  the  center  of  the 
circumscribed  sphere. 

708.  To  construct  a  sphere  of  given  radius  whose  surface  shall  contain 
three  given  points. 

709.  Also,  of  given  radius  whose  surface  shall  contain  two  given 
points  and  be  tangent  to  a  given  plane. 


Secs.  467,  468.] 


THE   SPHERE. 


339 


The  Relation  of  Spherical  Polygons  to  Polyhedral 

Angles. 

467.  If  the  center  of  a  sphere  is  at  the  vertex  of  a  pyramidal 
space,  the  pyramidal  surface  cuts  from  the  spherical  surface 
two  spherical  polygons. 


In  the  above  figure  the  two  polygons  are  ABCD,  A'B'C'D'. 

These  polygons  have  their  like-lettered  angles  and  sides 
equal  respectively. 

For  example.  /.  A  =  Z  A',  since  they  have  the  same  numerical  measure 
as  the  opposite  dihedral  angles  of  planes  ADOD'A'  and  ABOB'A'.  Also, 
AB  =  A'B\  since  the  central  angles  BOA  and  B'OA'  are  equal. 

468.  But  the  equal  elements  of  these  polygons  are  arranged 
in  reverse  order.  And  as  the  polyhedral  angles  are  called 
opposite  and  are  proved  (VI,  prop.  XXVI)  symmetric,  so  the 
spherical  polygons  are  called  opposite  spherical  polygons.  And 
since  these  have  just  been  shown  to  have  their  corresponding 
elements  equal  but  arranged  in  reverse  order,  they  are  called 
symmetric  spherical  polygons. 

Thus  all  opposite  polygons  are  symmetric ;  but  since  polygons  can 
slide  around  on  the  sphere,  it  follows  that  symmetric  polygons  are  not 
necessarily  opposite,  although  they  are  congruent  to  opposite  polygons. 


340  SOLID   GEOMETRY.  [Bk.  VIII. 

469.  Since  the  dihedral  angles  of  the  polyhedral  angles  have 
the  same  numerical  measure  as  the  angles  of  the  spherical 
polygons,  and  the  face  angles  of  the  former  have  the  same 
numerical  measure  as  the  sides  of  the  latter,  it  is  evident  that 
to  each  property  of  a  polyhedral  angle  corresponds  a  reciprocal 
property  of  a  spherical  polygon,  and  vice  versa.  This  relation 
appears  by  making  the  following  substitutions  : 

Polyhedral  Angles.  Spherical  Polygons. 

a.  Vertex.  a.  Center  of  Sphere. 

b.  Edges.  b.   Vertices  of  Polygon. 

c.  Dihedral  Angles.  c.  Angles  of  Polygon. 

d.  Pace  Angles.  d.  Sides. 

470.  In  addition  to  the  correspondences  between  poPyhedral 
angles  and  spherical  polygons,  it  will  be  observed  that  a 
relation  exists  between  a  straight  line  in  a  plane  and  a  great- 
circle  arc  on  a  sphere.  Thus,  to  a  plane  triangle  corresponds 
a  spherical  triangle,  to  a  straight  line  perpendicular  to  a 
straight  line  corresponds  a  great-circle  arc  perpendicular  to  a 
great-circle  arc,  etc.  The  word  arc  is  always  understood  to 
mean  great-circle  arc,  in  the  geometry  of  the  sphere,  unless  the 
contrary  is  stated. 

It  is  very  desirable  that  every  school  have  a  spherical  blackboard,  with 
large  wooden  compasses  for  the  drawing  of  both  great  and  small  circles. 
It  is  only  by  the  use  of  such  helps  that  students  come  to  a  clear  knowl- 
edge of  spherical  geometry.  If  such  a  blackboard  is  at  hand,  it  is  recom- 
mended that  many  problems  and  exercises  of  Book  I  be  investigated  on 
the  sphere.  E.g.  the  problem,  To  bisect  a  given  arc,  corresponds  to 
I,  prop.  XXXI,  and  the  solutions  are  quite  similar.  Likewise  the  prob- 
lems, To  bisect  a  given  angle,  To  draw  a  perpendicular  to  a  given  line 
from  a  given  internal  point,  etc.,  have  their  corresponding  problems  in 
spherical  geometry. 


Exercise.  710.  State  without  proof  the  proposition  in  the  geometry 
of  the  sphere  corresponding  to  the  following  :  Every  face  angle  of  a  con- 
vex polyhedral  angle  is  less  than  a  straight  angle. 


Props.  XI-XIII. 


THE   SPHERE. 


341 


Proposition  XI. 


471.    Theorem. 

(a)  In  any  trihedral  angle, 
each  face  angle  being  less 
than  a  straight  angle,  the 
sum  of  any  tic o  face  angles  is 


(a')  In  any  spherical  tri- 
angle, each  side  being  less 
than  a  semicircumference, 
the  sum  of  any  two  sides  is 


greater,  and  their  difference     greater,  and  their  difference 
less,  than  the  third  angle.  less,  than  the  third  side. 

Proof.     In  VI,  prop.  XXVII,  with  its   cor.   1,   (a)  has  been 
proved.     Hence  (a')  is  also  proved.  §  469 


Proposition  XII. 

472.    Theorem. 

(a)  In  any  polyhedral  angle, 
each  face  angle  being  less  than 
a  straight  angle,  any  face  an- 
gle is  less  than  the  sum  of  the 
remaining  face  angles. 


(a')  In  any  spherical  poly- 
gon, each  side  being  less  than 
a  semicircumference,  any  side 
is  less  than  the  sum  of  the 
remaining  sides. 


Proof. 


In  VI,  prop.  XXVII,  cor.  2,  (a)  has  been  proved. 
Hence  (a')  is  also  proved.  §  469 


Proposition   XIII. 

473.    Theorem. 

(a)  In  any  convex  polyhe-  (a')  In  any  convex  spherical 
dral  angle  the  sum  of  the  face  polygon  the  sum  of  the  sides 
angles  is  less  than  a  perigon.     is  less  than  a  circumference. 

Proof.         In  VI,  prop.  XXVIII,  (a)  has  been  proved. 

Hence  (a')  is  also  proved.  §  469 


342 


-SOLID   GEOMETRY. 


[Bk.  VIII. 


Proposition  XIV. 

474.  Theorem. 

(a)  No  face  angle  of  a  convex  (a')  No  side  of  a  convex 
p>olyhedral  angle  is  greater  spherical  polygon  is  greater 
than  a  straight  angle.  than  a  semicircumference. 

Proof.         From  §  461,  (a')  is  true. 

Hence  (a)  is  also  proved.  §  469 

475.  Definitions.  If  ABC  If  O-ABC  is  a  trihedral 
is  a  spherical  triangle,  and  A',      angle,  and  OA',  OB',  OC  are 


z? ^r 


B',  C  are  the  poles  of  a,  b,  c, 
respectively,  and  if  A  and  A', 
B  and  B',  C  and  C  lie  on  the 
same  side  of  a,  b,  c,  respec- 
tively, then  A  A'B'C  is  called 
the  polar  triangle  of  ABC. 


perpendiculars  to  a,  b,  c,  the 
faces  opposite  A,  B,  C,  respec- 
tively, and  if  A  and  A',  B  and 
B',  C  and  C  lie  on  the  same 
side  of  a,  b,  c,  respectively, 
then  trihedral  Z  O-A'B'C  is 
called  the  polar  trihedral  angle 
of  O-ABC. 


In  referring  to  polar  triangles  ABC,  A'B'C',  the  above  arrangement 
of  elements  will  always  be  intended.  Also,  in  referring  to  symmetric 
spherical  triangles,  ABC  and  A'B'C,  it  will  always  be  understood  that 
Z.A  =  /.  A',  etc.,  and  AB  —  A'B' ,  etc. 

The  polar  triangle  of  ABC  is  often  called  the  polar  of  ABC. 

It  is  evident  from  the  one-to-one  correspondence  of  §  475,  that  to 
every  proposition  concerning  polar  triangles  corresponds  a  proposition 
concerning  polar  trihedral  angles,  and  vice  versa. 


Prop.  XV.]  THE  SPHERE.  343 

PROrOSTTTOX    XV. 

476.   Theorem.     If  one  spherical  triangle  is  the  polar  of  a 
second,  then  the  second  is  also  the  polar  of  the  first. 


Given         a  spherical  triangle,  ABC,  and  A'B'C  its  pol^r. 
To  prove    that  A  ABC  is  the  polar  of  A  A'B'C. 
Proof.    1.  In  the  figure  suppose  AC,  AB',  drawn. 

2.  Then,  v  B1  is  a  pole  of b, 

.'.  AB'  is  a  quadrant.  Prop.  VI,  cor.  2 

Similarly,  v  C  is  a  pole  of  c, 
.'.  AC  is  a  quadrant. 

3.  .'.A  is  a  pole  of  a'.  Prop.  VII 
Similarly,  B  and  C  are  poles  of  J'  and  c',  respectively. 

4.  And  v  A,  A'  are  on  the  same  side  of  a',  and  so 
for  the  other  vertices  and  sides, 

.'.A  ABC  is  the  polar  of  A  A'B'C.  §  475 

Corollary.  If  one  trihedral  angle  is  the  polar  of  a  second, 
then  the  second  is  also  the  polar  of  the  first. 

For  from  the  one-to-one  correspondence  of  §  475,  the  proof  is  evi- 
dently identical  with  the  above. 

Note.  One  triangle  may  fall  entirely  within  or  entirely  without  its 
polar ;  or  one  may  be  partly  within  and  partly  without  the  other.  Simi- 
larly, one  trihedral  angle  may  fall  entirely  within  or  entirely  without  its 
polar  trihedral  angle,  or  may  be  partly  within  and  partly  without  the  latter. 


344  SOLID   GEOMETRY.  [Bk.  VIII. 

Proposition  XVI. 

477.  Theorem.  Any  angle  of  a  spherical  triangle  has  the 
same  numerical  measure  as  the  supplement  of  the  opposite 
side  of  its  polar. 


Given         ABC,  a  spherical  triangle,  and  A'B'C  its  polar. 

To  prove    that    the    numerical   measure   of   any    angle    C   is 
180° -c';    of  6",  180°  -  c. 

Proof.    1.  Suppose  a,  b  to  cut  c'  in  E\  D\  respectively,  and  a\ 
V  to  cut  c  in  E,  D,  respectively. 

2.  Measure  of  Z  C  =  that  of  tTe\  §  458,  2 
But  D%  =  AE'  +  IVB'  -  AB< 

=  90o4-90o-i7fe'  =  180°-c'.    Why  90°? 

3.  Similarly  for  Z  C,  substituting  A,  B,  D,  E,   for 
A,  B',  D\  E\  and  vice  versa,  in  the  above  proof. 

Corollaries.  1.  If  two  spherical  triangles  arc  mutually 
equiangular,  their  polars  are  mat  mi  11;/  equilateral ;  if  mutually 
equilateral,  their  polars  arc  mutually  equiangular. 

2.   The  sum  of  the  angles  of         2'.   The  sum  of  the  dihedral 

a  spherical  triangle  is  greater  angles  of  a  trihedral  angle  is 

than  one  and  less  than  three  greater  than  one  and  less  than 

straight  angles.  three  straight  angles. 


Prop.  XVII. 


THE   SPHERE. 


345 


For  by  prop.  XIII  (a'),     0  <  a'  +  b'  +  c'  <  360°. 
.-.  by  subtracting  from  3  ■  180°, 

3  •  180°>  (180°-  a')  +  (180°-  b')  +  (180°-  c')  >  180c 
.-.  by  prop.  XVI,  3  ■  180° >Z  A  +  Z  B  +  ZC  >180c 


478.  Definitions.    If  ABCD 

A'  is  a  spherical  polygon, 

and  A\  B',  C,  D',  are  the 

poles   of   XA,  AB,   BC,    CD, 

,   respectively,   and   if  A', 

B',  lie  on  the  same  side 

of    XA,    AB,    that    the 

polygon  does,  then  A'B'C'D' 


If  O-ABCD X  is  a  poly- 
hedral A,  and  OA',  OB',  OC, 

OD', are  _b  to  planes  OXA, 

OAB,  OBC, ,  respectively, 

and  if  A',  B', lie  on  the  same 

side   of  planes    OXA,    OAB, 

OBC, that  the  polyhedral 

angle  does,   then    O-A'B'C'D' 


is  called  the  polar  polygon      is  called  the   polar  poly- 

of  ABCD hedral  angle  of  O-ABCD 

Polar  trihedral  angles  are  also  called  supplemental  trihedral  angles. 

479.    A  spherical  triangle  is  said  to  be  birectangular  if  it  has 
two  right  angles,  trirectangular  if  it  has  three. 


Proposition  XVII. 
480.    Theorem. 

(a)  Two  opposite  or  two 
symmetric  trihedral  angles 
are  congruent  if  each  has 
two  equal  dihedral  angles,  or 
two  equal  face  angles. 


(a')  Two  opposite  or  two 
symmetric  spherical  triangles 
are  congruent  if  each  has  tiro 
equal  angles  or  two  equal 
sides. 


Proof  for  (a').    1.  Their  sides  and  angles  are  respectively  equal 
but  arranged  in  reverse  order.  §  468 

2.  But  if  to  the  order  ABC  corresponds  B'A'C, 

and  if  B1  =  A',  then  B'  and  A'  may  be  interchanged. 

3.  Then  to  the  order  ABC  will  correspond  A'B'C, 
and  the  A  are  congruent  by  superposition. 


346  SOLID    GEOMETRY.  [Bk.  VIIL 

Proposition   XVIII. 

481.    Theorem.     Tivo  symmetric  spherical  triangles  on  the 
same  sphere  or  on  equal  spheres  are  equal. 


Given         two  symmetric  spherical  triangles,  ABC,  A'B'C,  on 
the  same  sphere. 

To  prove    that  A  ABC  =  A  A'B'C. 

Proof.    1.  The  plane  of  A,  B,  C  determines  a  small  circle. 

2.  Let  0  be  the  pole  of  the  O,  and  similarly  for  0'  and 
spherical  A  A'B'C. 

3.  Then  V  side  AB  =  side  A!B\  .'.  chord  AB  =  chord 
A'B'.  (In  the  figure  they  are  not  drawn  because 
AB  is  so  nearly  straight.)  Ill,  prog.  1 1  L 
Similarly  for  chords  BC,  B'C,  and  CA,  CA'. 

4.  .'.  plane  A  ABC  ^  plane  A  A'B'C.         I,  prop.  XII 

5.  .'.  O  ABC  —  O  A'B'C,  being  circumscribed  about, 
congruent  plane  A.  Why  ? 

6.  .-.  0?1  =  0?i  =  6C  =  OA'  =  (fji'  =  CC'.       Why  ? 

7.  .".  spherical  A  JO/7  £  .-l'O'T?',  BOC  £  5'0'C, 
C04  ^  C"0'4'.  §§  468,  480  (a') 

8.  .'.  A  .!/;(/  =  A  A'B'C.  Ax.  2 


Prop.  XIX.]  THE  SPFfERE.  347 

Proposition  XIX. 

482.    Theorem.      Two  triangles  on  the  same  sphere,  or  on 
equal  spheres,  are  either  congruent  or  symmetric  and  equal  if 

(a)   two  sides  and   the   in-       (b)   two  angles  and  the  in- 
cluded angle  eluded  side 
of  the  one  figure  are  equal  to  the  corresponding  parts  of  the 
other. 


Proof.  If  the  parts  are  arranged  in  the  same  order,  the 
triangles  can  be  brought  into  coincidence,  as  in  I, 
props.  I,  II. 

If  they  are  arranged  in  reverse  order,  then  one 
triangle  is  congruent  to  the  triangle  symmetric  to 
the  other.  Why? 

Corollary.  Two  triliedral  angles  are  either  congruent  or 
symmetric  and  equal  if 

(a)  two  face  angles  and  the      (b)  two  dihedral  angles  and  the 
included  dihedral  angle  included  face  angle 

of  the  one  figure  are  equal  to  the  corresponding  parts  of  the 
other. 

For  from  the  one-to-one  correspondence  of  §  475,  the  proof  is 
evidently  identical  with  the  above,  without  the  labor  of  drawing  the 
figures. 


348  SOLID   GEOMETRY.  [Bk.  VIII. 

Proposition  XX. 
483.    Theorem. 

(a)  If  a  trihedral  angle  has  (a')  If  a  spherical  triangle 
two  dihedral  angles  equal  to  has  two  angles  equal  to  each 
each  other,  the  opposite  face  other,  the  opposite  sides  are 
angles  are  equal.  equal. 

Given         the  A  ABC,  with  ZA  =  ZB. 

To  prove    that  a  =  b. 

Proof.    1.  Let   A  A'B'C    be    symmetric    to   A  ABC,    so   that 
a  =  a',  b  =  b',  etc. 

2.  Then  v  ZA  =  ZB, 

.'.  ZA'  must  equal  ZB',  and  the  A  are  congruent 
and  a'  =  b.  Prop.  XIX 

3.  But  '•'  a  =  a\  and  a'  =  b, 

.'.  a  =  b,  which  proves  (a').  Ax.  1 

Hence  (a)  is  also  proved.  §  469 

Corollaries. 

1.  (a)  If  a  trihedral   angle  (a')   ^4??  equiangular  spher- 
has  its  three  dihedral  angles      ical  triangle  is  equilateral, 
equal,   it    has    also    its    three 

face  angles  equal. 

In  A  ABC,  if  Z  A  =  Z  B,  then  a  =  b  ;  and  if  Z  C  also  equals  Z  B, 
c  also  equals  b. 

.:  if  Z  A  =  Z  B  =  Z  C,  a  =  b  =  c.  Ax.  1 

2.  (a)  If  a  trihedral  angle  (a')  If  a  spherical  triangle 
has  two  face  angles  equal  to  has  two  sides  equal  to  each 
each  other,  the  opposite  dihe-  other,  the  opposite  angles  are 
dral  angles  are  equal.  equal. 

The  proof  is  almost  identical 
with  that  of  I,  prop.  Ill,  and 
hence  is  left  for  the  student. 


Prop.  XXL] 


THE   SPUE  UK. 


349 


Proposition  XXI. 

484.    Theorem.     Two  triangles  on  the  same  sphere,  or  on 
equal  spheres,  are  either  congruent  or  symmetric  and  equal  if 

(a)  the  three  sides  (b)  the  three  angles 

of  the  one  figure  are  equal  to  the  corresponding  parts  of  the 
other. 


(a)  Given  A  ABC,  A'B'C,  mutually  equilateral,  the  sides  being 
arranged  in  the  same  order  ;  also  A  ACX  symmetric 
to  A  A'B'C. 

To  prove  that  A  ABC  =  A  A'B'C,  A  ABC  is  symmetric  to 
A  ACX. 

Proof.    1.  Place  A  ACX  as  in  the  figure  ;   draw  BX. 

Then  Z  BXC  =  Z  CBX, 

and  Z  AXB  =  Z  XBA.        Prop.  XX,  cor.  2 


2.  .\ZAXC  =  ZCBA, 
i.e.  ZB  =ZX  =  ZB'. 

Similarly  with  the  other  angles. 

3.  .\  A  A'B'C  ^  A  ABC. 

4.  And  A  ABC  is  symmetric  to  A  ACX 


Ax.  3 

Why  ? 
Why  t 


350  SOLID    GEOMETRY.  [Bk.  VIII. 

(b)  Given  (Let  the  student  state  it.) 


To  prove  (Let  the  student  state  it.) 

Proof.    1.  Their  polars  are  mutually  equilateral.  Why  ? 

2.  .'.  their  polars  are  congruent  or  symmetric.      Why  ? 

3.  .*.  A  ABC  and  A  A'B'C  are  mutually  equilateral. 

Props.  XV,  XVI,  cor.  1 

4.  .'.AABC*£  or  symmetric  to  A  A'B' 6".  Prop.  XXI  (a) 

Corollary.      Two  trihedral  angles  are  either  congruent  or 
symmetric  and  equal  if 

(a)  the  three  face  angles  (b)  the  three  dihedral  angles 

of  the  one  figure  are  equal  to  the  corresponding  parts  of  the  other. 


Exercises.  711.  A  plane  isosceles  triangle  can  have  its  equal  sides  of 
any  length.    Discuss  as  to  a  spherical  isosceles  triangle  on  a  given  sphere. 

712.  As  with  plane  triangles,  the  pole  (circuracenter)  may  fall  outside 
the  triangle,  or  on  a  side.     Prove  theorem  481  for  those  cases. 

713.  Prove  I,  prop.  XII  (a  corresponding  theorem  of  Plane  Geometry) 
by  the  method  of  prop.  XVIII. 

714.  Draw  the  figure  of  a  spherical  quadrilateral  and  its  polar;  also 
of  a  four-faced  polyhedral  angle  and  its  polar. 

715.  Prove  that  if  one  spherical  polygon  is  the  polar  of  another,  then 
the  second  is  the  polar  of  the  first.  State  the  reciprocal  theorem  for 
polyhedral  angles.     (The  special  case  of  the  quadrilateral  may  be  taken.) 


Prop.  XXII.] 


THE   SPHERE. 


351 


Proposition   XXII. 

485.  Theorem.  For  a  great  circle  to  be  perpendicular  to  a 
small  circle,  it  is  necessary  and  it  is  sufficient  that  the  circum- 
ference of  the  former  pass  through  a  pole  of  the  latter. 


Given         a  small  circle  S,  with  P  and  P'  its  poles,  G  a  great 
circle,  and  0  the  center  of  the  sphere. 

To  prove    that  for   G  to  be  J_  to  S  it  is  necessary  and  it  is 
sufficient  that  its  circumference  pass  through  P. 

Proof.    1.  PP'  _L  8.  Def.  pole 

2.  And  if   G  passes  through  P  it  passes  through  PP', 
and  G  ±  S.  VI,  prop.  XVIII 

3.  .*.  it  is  sufficient  that  G  contain  P. 

4.  Furthermore  it  is  necessary;  for  if  G  _L  S,  then  PP 
lies  in  G  or  else  PP'  II  G.  Why  ? 

5.  But  PP'  is  not  II  to  G,  for  each  contains  0.     Why  ? 

6.  .'.  it  is  necessary,  and  it  is  sufficient,  that  6r  contain  P. 

Corollaries.  1.  Through  a  poi?it  X,  within  or  on  a,  sphere, 
it  is  possible  to  pass  one  great  circle  perpendicular  to  a  given 
circle  S,  and  only  one  unless  X  is  on  the  straight  line  through 
the  poles  of  S. 

For  PP'  passes  through  the  center  0,  and  PP'  and  X  determine  a  great 
circle  _L  S,  unless  X  is  on  PP'.    (Why  ?)    May  X  be  without  the  sphere  ? 


352 


SOLID   GEOMETRY. 


[Bk.  VIII. 


2.  If  the  circumferences  of  two  great  circles  are  drawn  per- 
pendicular to  a  third  circumference,  they  will  intersect  at  the 
poles  of  the  circle  of  that  third  circumference. 

486.  Definition.  If  a  great  circle  is  perpendicular  to  a  small 
circle,  their  circumferences  are  said  to  be  perpendicular  to  each 
other. 


Proposition  XXIII. 

487.  Theorem.  If  from  a  point  on  a  sphere  arcs  of  great 
circles  both  perpendicular  and  oblique,  are  drawn  to  any 
circumference,  then, 

1.  The  shorter  perpendicular  is  less  than  any  oblique; 

2.  Two  obliques  cutting  off  equal  arcs  from  the  foot 
of  this  perpendicular  are  equal ; 

3.  Of  two  obliques  cutting  off  unequal  arcs  from  the 
foot  of  this  perpendicular,  the  one  cutting  off  the  greater 
arc  is  the  greater. 


Given  S,  any  circle  of  a  sphere ;  P  any  point  on  the 
spherical  surface;  minor  arcs,  PC  J_  circumference 
S,  PA,  PB,  PD  obliques  ;  PC  =  CD,  and  AC  >  CD 
or  its  equal  BC. 

To  prove    that  (1)  PC  <  PB ; 

(2)  PB  =  PD ; 

(3)  Rk  >  PD  or  its  equal  PB. 


Prop.  XXIII.]  THE  SPHERE.  353 

Proof.    1.  Suppose  N  the  pole  of  S,  on  the  same  side  of  S  as 
P;  draw  NA,  NB,  ND. 

2.  CP  produced  passes  through  X.    Prop.  XXII,  cor.  2 

3.  NB,  or  its  equal  NC,  <  NT  +  PB.     Prop.  XI,  (a') 

4.  .'.PC<PB.  Why? 

5.  The  radius  of  O  S  through  C  is  _L  to  chord  PD  and 
bisects  it  as  at  Q  (not  shown  in  figure).  Why  ? 

6.  .'.I)B±  plane  iVP C.  Why  ? 

7.  .\  plane  ABQP  =  PQB,  Why? 
and  plane  A  #PD  ^  ()PP.                  I,  prop.  I 

8.  .\PB  =  PB.  Ill,  prop.  IV 

9.  Pi7  +  EB  >  PB,  and  pQ  >  7^P.  Why  ? 
10.               .'.  PE+  PA,  or  PA,  >  PB. 

488.  Definition.  The  excess  of  the  sum  of  the  angles  of 
a  spherical  n-gon  over  (ii  —  2)  straight  angles  is  called  the 
spherical  excess  of  the  ?i-gon. 

Hence  the  spherical  excess  of  a  2-gon  (lime),  3-gon  (triangle),  4-gon, 
is  the  excess  of  the  sum  of  its  angles  over  0,  1,  2, straight  angles. 


Exercises.  716.  Prove  that  if  in  prop.  XVI  the  word  polygon  is  sub- 
stituted for  triangle,  the  resulting  theorem  is  true,  and  state  the  corollary 
that  follows  from  it,  analogous  to  corollary  1  of  prop.  XVI. 

717.  What  is  meant  by  the  spherical  excess  of  a  spherical  decagon  ? 
What  is  the  spherical  excess,  in  degrees,  of  a  triangle  whose  angles  are 
75°,  90°,  100°? 

718.  What  is  the  spherical  excess,  in  radians,  of  a  triangle  whose 
angles  are  80°,  90°,  100°?  Also  of  a  triangle  whose  angles  are  1,  2,  and 
3  radians,  respectively  ? 


354  SOLID    GEOMETRY.  [Bk.  VIII. 

Proposition  XXIV. 

489.    Theorem.     A  spherical  triangle  equals  a  lune  icJwse 
angle  is  half  the  spherical  excess  of  the  triangle. 


B' 

Given  T,  a  spherical  triangle,  with  angles  a,  b,  c. 

To  prove    that  T  =  a  lime  whose  angle  is  \  (a  +  b  -f  c  —  st.  Z). 

Proof.  1.  Let  A,  B,  C  =  limes  of  A  a,  b,  c,  respectively  (in 
the  figure  they  are  AA',  BB',  CC),  and  S=  surface 
of  sphere. 

2.  A  AB'C  and  ABC  are  mutually  equilateral,  for 
AC  +  CA  =  semicircumference  =  AC  +  CA  ;  hence 
AC  =  AC,  and  so  for  the  other  sides.  Ax.  3 

3.  .'.  A  AB'C  =  A  ABC,  so  that  T  +  A  AB'C  =  hme  A. 

Prop.  XXI 

4.  .\A  +  (B-  T)  +  (C-T)  =$S,  Ax.  8 
or          T=±(A  +  B  +  C-iS).                 Axs.  3,  7 

5.  But  '.*  i  S  =  a  lime  whose  Z  is  a  st.  Z,  §  402 
.'.  T  =  a  lune  whose  Z  is  J-  (a  +  />  +  c  —  st.  Z). 

Corollary.  A  spherical  polygon  equals  a  lune  whose  angle 
is  half  the  spherical  excess  of  the  polygon. 

For  the  polygon  can  he  cut  into  triangles  as  in  Plane  Geometry. 

The  practical  method  of  measuring  a  spherical  polygon  is  given  in 
§  493,  cor.  3. 


Prop.  XXV.]     THE  MENSURATION  OF    THE   SPHERE.       355 
4.     THE   MENSURATION   OF   THE    SPHERE. 

Proposition  XXV. 

490.    Theorem.      The  area   of  the  surface  of  a  sphere  of 
radius  r  is  -I  7rr2. 

B 


A'    M'    B'O 


Proof.    1.  A  semicircle  cut  off  by  a  diameter  XX,  revolving 
about  X'X  as  an  axis,  generates  a  sphere. 

2.  Let  AB  be  one  of  a  number  of  chords  inscribed  in 
arc  XBX',  forming  half  of  a  regular  polygon. 

Let  031  ±  AB,  thus  bisecting  it ;  III,  prop.V 

let  A  A',  MM,  BB',  all  be  _L  to  X'X,  and  AC  II  X'X. 

3.  Then  AB,  revolving  about  axis  X'X,  generates  the 
surface  I  =  2  it  -  AB  ■  M'M  Prop.  Ill,  cor.  2 

4.  But  V  A  A  CB  —  A  J/J/'  0,  Why  ? 

.-.  03f  :  J/' J/  =  AB :  J  C  =  A B  :  .!'£'. 

5.  .'.  ^LB  •  IT If  =  A'B'  •  OK  IV.  pr0p.  I 

6.  .'.  Z  =  2it  •  .!'£'  •  OM.  Subst.  in  3 

7.  Summing  for  all  frustums,  including  two  cones,  the 
sum  of  the  lateral  surfaces  =  2  7r  •  OM  •  (X'A1  + 
A'B'  + )  =  2tt  •  OM -2  r.  Axs.  2,  8 

8.  But  if  the  number  of  sides  increases  indefinitely,  the 
sum  of  the  lateral  surfaces  ==  surface  of  sphere,  s, 
and  OM  =  r ; 

.'.  *  =  2ir  •  r  •  2r  =  4  7rr\  IV,  prop.  IX,  cor.  1 


356  SOLID    GEOMETRY.  [Bk.  VIII. 

491.  Definitions.  That  part  of  a  spherical  surface  which 
is  included  between  two  parallel  planes  which  cut  or  touch 
the  surface,  is  called  a 

zone.  / —  — \ 

The  solid  bounded 
by  the  zone  and  the 
two  parallel  planes  is 

x  r  Zones  and  spherical  segments.    In  first  figure, 

Called    a   Spherical   seg-  lower  base  is  zero. 

ment. 

The  distance  between  the  two  parallel  planes  determining  a 
zone  and  a  spherical  segment  is  called  the  altitude  of  the  zone 
and  the  segment. 

The  circumferences  in  which  the  planes  intersect  the  spher- 
ical surface  are  called  the  bases  of  the  zone,  and  the  circles  are 
called  the  bases  of  the  segment. 

In  case  of  tangent  planes  the  bases  may  one  or  both  reduce  to  zero. 
If  one  base  only  reduces  to  zero,  the  zone,  or  segment,  is  said  to  have 
one  base. 

492.  Definition.  As  a  plane '  angle  is  often  said  to  be 
measured  by  the  ratio  of  the  intercepted  arc  to  the  whole 
circumference  (§  256),  so  a  polyhedral  angle  is  said  to  be 
measured  by  the  ratio  of  the  intercepted  spherical  polygon  to 
the  whole  spherical  surface. 

The  practical  method  of  measuring  a  polyhedral  angle  is  given  in 
§  493,  cor.  4. 

493.  Corollaries.  1.  The  area  of  a  zone  of  altitude  a,  on 
a  sphere  of  radius  r,  is  2  7rra. 

For,  prop.  XXV,  step  7,  the  sum  of  the  lateral  surfaces  may  approach 
zs  their  limit  a  zone,  in  which  case  X'A'  +  A'B'  + =  a,  and  OM  ==  r. 

2.  The  area  of  a  lune  of  angle  a  (expressed  in  radian  measure) 
on  a  sphere  of  radius  r,  is  2  a?2. 

By  prop.  X,  I :  4  7tr2  =  a  :  2  it. 

3.  The  area  of  a  spherical  polygon  of  spherical  excess  a 
(expressed  in  radian  measure)  is  av2. 


Prop.  XXV.]    THE  MENSURATION   OF   THE  SPHERE.         357 

For  by  prop.  XXIV,  the  polygon  equals  the  lune  whose  angle  is  a/2. 
.-.  the  area  =  2  •  —  •  r-  =  ar2,  by  cor.  2. 

4.   The  measure   of  a  polyhedral   angle   whose  intercepted 
spherical  polygon  has  a  spherical  excess   a  is 

For  by  definition,  §  490,  it  is 


a 

4-7T 

ar'2 
4  7tr2 


5.  The  area  generated  by  a  chord  of  a  circle  revolving  about 
a  diameter  which  does  not  cut  it,  equals  2tt  times  the  product 
of  its  projection  on  that  diameter,  and  the  distance  from  the 
center  to  the  chord.      (Why  ?) 

G.   The  areas  of  two  spheres  are  propjortional  to  the  squares 

of  their  radii. 

_      a       4  7T/-2       r°- 

For  —  = =  — ■ 

a       4  tit1      r 2 


Exercises.  719.  What  is  the  area  of  a  spherical  triangle  the  sura  of 
whose  angles  is  4  radians,  on  a  sphere  of  radius  1  ft.  ? 

720.  Also  of  one  the  sum  of  whose  angles  is  270°,  r  being  2  ft.  ? 

721.  Also  of  one  the  sum  of  whose  angles  is  180°,  on  any  sphere  ? 

722.  Also  of  one  the  sum  of  whose  angles  is  237°  29',  r  being  10  in.  ? 

723.  Also  of  a  spherical  quadrilateral  the  sum  of  whose  angles  is 
417°  29',  on  a  sphere  of  radius  2  in.  ? 

724.  Also  of  a  spherical  pentagon  the  sum  of  whose  angles  is  4  straight 
angles,  on  a  sphere  of  radius  5  in.  ? 

725.  What  is  the  measure  in  radians  of  a  polyhedral  angle  the  spherical 
excess  of  whose  intercepted  spherical  polygon  is  8  it  ? 

726.  What  is  the  ratio  of  a  trihedral  angle  the  sum  of  the  angles  of 
whose  intercepted  spherical  triangle  is  1.5  7r  radians,  to  a  tetrahedral 
angle  the  sum  of  the  angles  of  whose  intercepted  spherical  quadrilateral 
is  2.5  it  radians  ? 

727.  What  is  the  area  of  a  spherical  triangle  whose  angles  are  70°, 
80°,  90°,  on  a  sphere  of  diameter  20  in.  ? 

728.  Show  that  a  trirectangular  triangle  is  its  own  polar. 

729.  The  locus  of  points  on  a  sphere,  from  which  great-circle  arcs 
perpendicular  to  the  arms  of  an  angle  are  equal,  is  the  great-circle  arc 
bisecting  that  angle. 


358 


SOLID    GEOMETRY. 


[Bk.  VIII. 


Proposition  XXVI. 

494.  Theorem.  Two  solids  lying  between  two  parallel 
planes,  and  such  that  the  ttvo  sections  made  by  any  plane 
parallel  to  the  given  planes  are  equal,  are  themselves  equal. 


Given         two  solids,   S,   S',    lying   between   parallel   planes, 
M,  N,  and  such  that  the  two  sections  A,  A\  or  B, 

B\ ,  made  by  any  plane  Q,  or  R, ,  are  equal, 

i.e.  A  =  A'B  =  B\ 


To  prove    that 


s=sr 


Proof.  1.  Let  K,  K'  be  two  segments  of  S,  S',  lying  between 
the  sections  A  and  B,  and  A'  and  B' ;  let  the  alti- 
tude of  K,  K'  be  1/n  of  the  altitude  h  of  S  and  S'. 

2.  Suppose  two  straight  lines  to  move  so  as  always  to 
be  perpendicular  to  Q,  and  to  touch  the  perimeters 
of  A,  A',  thus  generating  two  cylinders  (or  prisms, 
or  combinations  of  cylinders  and  prisms)  of  altitude 
h/n  as  in  Fig.  2.  As  the  volumes  of  both  prisms 
and  cylinders  are  expressed  by  the  same  formula, 
v  =  bh}  we  may  speak  of  these  solids  as  cylinders, 
C,  C. 

3.  Then  C  =  C,  since  they  have  equal  bases  and  alti- 
tudes ;  and  so  for  other  pairs  of  cylinders  described 

in  the  Same  wav.  with  altitude  h/n. 


Prop.  XXVI.]      THE  MENSURATION  OF  THE  SPHERE.       359 

4.  .'.  the  sum  of  the  solids  like  C  =  the  sum  of  the 
solids  like  C",  whatever  n  equals. 

5.  But  if  n  increases  indefinitely,  h/n  decreases  in- 
definitely, and  it  is  evident  that  the  sum  of  the 
solids  like  C  =  S,  while  the  sum  of  the  solids  like 
6"  =  S\ 

6.  .'.  S=  S'.  IV,  prop.  IX,  cor.  1 

495.  This  important  proposition  is  known  as  Cavalieri's 
theorem.  It  will  be  seen  that  VII,  prop.  XV,  is  merely  a 
special  case  of  this  proposition.  We  shall  base  the  mensura- 
tion of  the  volume  of  the  sphere  upon  it.  Solids  of  this  kind 
are  often  called  Cavalieri  bodies. 


Exercises.  730.  A  spherical  triangle  is  to  the  surface  of  the  sphere 
as  the  spherical  excess  is  to  eight  right  angles. 

731.  The  locus  of  points  on  a  sphere,  from  which  great-circle  arcs  to 
two  fixed  points  on  the  sphere  are  equal,  is  the  circumference  of  a  great 
circle  perpendicular  to  the  arc  joining  those  points  at  its  mid-point. 

732.  There  is  evidently  a  proposition  of  plane  geometry  analogous  to 
Cavalieri's  theorem,  beginning,  "Two  plane  surfaces  lying  between  two 
parallel  lines "     State  this  proposition  and  prove  it. 

733.  From  ex.  732  prove  that  triangles  having  equal  bases  and  equal 
altitudes  are  equal. 

734.  What  is  the  ratio  of  the  surface  of  a  sphere  to  the  entire  surface 
of  its  hemisphere  ? 

735.  Prove  that  the  areas  of  zones  on  equal  spheres  are  proportional 
to  their  altitudes. 

736.  Find  the  ratio  of  the  surfaces  of  two  spheres,  in  terms  of  their 
radii,  i\  and  r2. 

737.  What  is  the  ratio  of  the  area  of  a  great  circle  of  a  sphere  to  the 
area  of  its  spherical  surface  ? 

738.  If  a  meter  is  0.0000001  of  a  quadrant  of  the  earth's  circumfer- 
ence, and  the  earth  is  assumed  to  be  a  sphere,  how  many  square  myria- 
meters  of  surface  has  the  earth  ? 

739.  What  is  the  radius  of  the  sphere  whose  area  is  1  square  unit  ? 
Answer  to  0.001. 


360 


SOLID   GEOMETRY. 


[Bk.  VIII. 


Proposition  XXVII. 

496.    Theorem.     The  volume   of  a  sphere  of  radius  r  is 
expressed  by  the  formula  v  =  j  77-r3. 


Proof.  1.  Suppose  the  sphere  circumscribed  by  a  cylinder, 
and  suppose  two  cones  formed  with  the  bases  of 
the  cylinder  as  their  bases,  and  their  vertices  at  the 
center  of  the  sphere. 

Suppose  the  solid  to  be  cut  by  a  plane  Q,  parallel 
to  the  bases,  and  x  distant  from  the  center  of  the 
sphere. 

2.  Then  since  x  also  equals  the  radius  of  the  O  cut 
— ^    from  the  cone,  because  the  altitude    of   the    cone 
equals  the  radius  of  its  base, 
.*.  area   of    ring    CD   between   cone   and   cylinder 


=  7r(r2-a-2). 

3.  But  the  area  of  the  O  AB  cut  from  the  sphere  is 
also  it  (r2  —  x2),  because  its  radius  is  V r*  —  .>■'-. 

4.  .'.  the  sphere  and  the  difference  between  the  cone 
and  cylinder  are  two  Cavalieri  bodies,  and  .'.  they 
are  equal.  Prop.  XXVI 

5.  .'.  v  =  tt>-2  •  2r  -  Trr2  ■  ",'*  =  i  tt/-8.  Why  ? 


Prop.  XXVII.]      THE  MENSURATION  OF   THE  SPHERE.      361 


Corollaries.  1.  The  volume  of  a  sphere  equals  two- 
thirds  the  volume  of  the  circumscribed  cylinder.  (Arehimedes's 
theorem.) 


/ 


The  volume  of  the  circumscribed  cylinder  is  evidently  itr-  •  2  r,  or  2  7tr\ 


2.    The  volume  of  a  sphere  equals  the  product  of  its  surface 
by  one-third  of  its  radius. 


For  the  surface  is  -4  7rr'2,  prop.  XXV  ;  and  f  itf1 


r  •  4  7tr~. 


3.  The  volumes  of  two  spheres  are  proportional  to  the  cubes 

of  their  radii. 

4.  The  volume  of  a  spherical  segment  of  one  hasp,  of  altitude 
a,  is  expressed  by  the  formula  v  =  \  7ra2(3  r  —  a). 

For,  as  in  the  theorem,  it  equals  the  difference  between  a  circular 
cylinder  of  radius  r  and  altitude  a,  and  the  frustum  of  a  cone,  of  the 
same  altitude  and  with  bases  of  radii  r  and  (r  —  a). 

.-.  v  =  itr-a  —  \na  \r-  +  (r  —  a)'2  +  r  (r  —  a)]  Prop.  IV,  cor.  1 

=  i7ra2(3r-a). 

497.    Definitions.     A  spherical  sector  is  the  portion  of  a  sphere 
generated  by  the  revolution  of  a  circular 
sector  about  any  diameter  of  its  circle  as 
an  axis. 

The  base  of  the  spherical  sector  is  the 
zone  generated  by  the  arc  of  the  circular 
sector,  and  the  altitude  is  the  altitude  of 
that  zone. 

If  the  base  of  the  spherical  sector  is  a 
zone  of  one  base  only,  the  spherical  sector 
is  called  a  spherical  cone. 


Spherical  sectors.  The 
upper  one  a  spherical 
cone. 


362  SOLID   GEOMETRY.  [Bk.  VIII. 

Corollaries.  1.  The  volume  of  a  spherical  cone,  whose  base 
b  has  an  altitude  a,  is  expressed  by  the  formula  v  =  J  7rr2a,  or 
v  =  i  br. 

For  it  evidently  equals  the  sum  of  a  cone  and  a  spherical  segment 
of  one  base.  What  does  the  latter  equal,  by  §  496,  cor.  4  ?  Show  that 
the  cone  =  \  it  (r  —  a)  [r2  —  (r  —  a)2].  Then  add  the  results,  and  show 
that  the  sum  is  §  itr2a.     But  6  =  2  nra.     (Why  ?) 

2.  The  volume  of  a  spherical  sector,  whose  base  is  b  and  alti- 
tude a,  is  expressed  by  the  formula  v  =  §7rr2a,  or  v  =  ±  br. 

For  it  equals  the  difference  between  two  spherical  cones. 
Suppose  these  to  have  altitudes  a\ ,  «2 , 

and  bases       b{ ,  b2 , 

and  volumes  Vi ,  v2 ,  respectively. 
Then  v  =  Vi  —  v2  =  j  nr2a\  —  § 7tr2a2  —  §  nr2  (<3i  —  a2). 
But  ci!  -  a2  =  a.     (Why  ?) 
.-.  u  =  1 7rr2a.     Now  show  that  v  =  \br.- 



Exercises.  740.  Show  that  if  the  directrix  of  a  cylinder  is  the  cir- 
cumference of  a  great  circle  of  a  sphere,  and  the  generatrix  is  perpen- 
dicular to  that  circle,  and  the  bases  of  the  cylinder  are  circles  tangent 
to  the  sphere,  then  the  cylinder  may  be  said  to  be  circumscribed  about 
the  sphere. 

741.  After  considering  ex.  740,  show  that  the  surface  of  a  sphere  is 
two-thirds  the  entire  surface  of  the  circumscribed  cylinder.    (Archimedes. ) 

742.  Find  the  ratio  of  a  spherical  surface  to  the  cylindrical  surface  of 
the  circumscribed  cylinder. 

743.  What  is  the  radius  of  that  sphere  the  number  of  square  units  of 
whose  area  equals  the  number  of  linear  units  in  the  circumference  of  one 
of  its  great  circles  ? 

744.  What  is  the  ratio  of  the  entire  surface  of  a  cylinder  circum- 
scribed about  a  sphere  to  the  entire  surface  of  its  hemisphere  ? 

745.  What  is  the  area  of  the  entire  surface  of  a  spherical  segment  the 
radii  of  whose  bases  are  n,  r2,  the  radius  of  the  sphere  being  r  ? 

746.  A  cone  has  for  its  base  a  great  circle  of  a  sphere,  and  for  its 
vertex  a  pole  of  that  circle.  Find  the  ratio  of  the  curved  surfaces  of  the 
cone  and  hemisphere  ;  of  the  entire  surfaces. 

747.  Show  that  the  area  of  a  zone  of  one  base  (the  other  base  is  zero) 
equals  that  of  a  circle  whose  radius  is  the  chord  of  the  generating  arc. 


Prop.  XXVIII.]    THE  MENSURATION  OF  THE  SPHERE.      3G3 

Proposition  XXVIII. 

498.  Theorem.  The  volume  of  a  spherical  segment  of 
altitude  a,  whose  bases  have  radii  rv  r2,  is  expressed  by 
the  formula 

v  =  i  vra  [3  (i-!2  +  r22)  +  a2]  ,  or 

v  =  1 7ra  (i*!2  +  r22)  +  §  7ra3. 


Proof. 


1.  Let  the  above  figure  represent  a  segment  cut  from 
the  figure  of  prop.  XXVII. 

2.  Then  if  the  distances  of  the  circles  of  radii  i\  and 
r2,  from  the  center  0,  are  xx  and  x2,  respectively, 
the  radii  of  the  bases  of  the  frustum  of  the  cone 
are  xx  and  x2.  ^Vhy  ? 

3.  v  =  cylinder  —  frustum,  Prop.  XXVII,  step  4 

=  iri^a  —  i  Tra  (xt*  +  x2  +  xlx2)    Prop.  IV,  cors.  1,  6 

=  i  ira  (6  >»2  -  2  a^2  -  2  a^  -  2  ay2) 

=  i  tt«  [3  (>-2  -  x,2)  +  3  (r2  -  x22)  +  («,  -  z2)2]. 

4.  But  '•'  <x  =  xx  —  x2,  and  rx2  =  i2  —  a-!2, 
and     r2  =  ^  —  x22, 

.'.v  =  ±7ra[3(rl2  +  r22)  +  a2] 
=  i7ra(r12  +  7-22)  +  ^7ra3. 


Exercise.  748.  Within  an  equilateral  triangle  of  side  s  is  inscribed 
a  circle  ;  the  triangle  revolves  about  one  of  its  axes  of  symmetry,  thus 
generating  a  sphere  and  a  cone.     Find  the  ratio  of  their  curved  surfaces. 

749.    Also  find  the  ratio  of  their  entire  surfaces. 


3G4 


SOLID   GEOMETRY 


[Bk.  VIII. 


5.     SIMILAR   SOLIDS. 

499.  The  definitions  of  similar  systems  of  points  and  similar 
figures  given  in  §§  257,  258  are  not  limited  to  plane  figures. 
The  lines  forming  the  pencil  need  not  be  coplanar.  In  case 
they  are  not  coplanar,  a  pencil  of  lines  is  often  called  a  sheaf 
of  lines.  The  similar  figures  may  then  be  plane,  or  they  may 
be  curved  surfaces,  or  solids,  etc.  The  definitions  and  corol- 
laries on  pages  182-184  are  therefore  the  same  for  solid 
figures  as  for  plane,  and  should  be  reviewed  as  part  of  this 
section. 

Polyhedra  which  have  equal  face  and  equal  dihedral  angles, 
and  equal  edges,  but  have  these  parts  arranged  in  reverse 
order,  are  said  to  be  symmetric. 

The  polyhedral  angles  are  then  respectively  symmetric. 


Proposition  XXIX. 

500.  Theorem.  If  two  polyhedra  are  similar,  their  corre- 
sponding face  and  dihedral  angles  are  equal,  their  correspond- 
ing p>olyhedral  angles  are  either  congruent  or  symmetric,  and 
their  corresponding  edges  are  in  proportion,  the  constant 
ratio  being  the  ratio  of  similitude. 

D, 


Given         two  similar  polyhedra,  JlBlCl and    A.Ji.J\ , 

or  AXBXCX and  ASB9CB 


Prop.  XXIX.]  SIMILAR   SOLIDS.  365 

To  prove    that  (1)  Z  BlA1I)1  =  Z  B%AJ>%  or  Z  B3AZD3 ; 

(2)  dihedral  angle  with  edge  AXBX  =  dihedral  angle 
with  edge  A2B2,  or  with  edge  A3B3; 

(3)  polyhedral  angle  Ax  =  polyhedral  angle  A2  and 
is  symmetric  to  polyhedral  angle  A3  as  arranged  in 
the  figure ;  and 

(4)  AXBX  :  A2B2  =  the  ratio  of  similitude. 

Proof.    1.  Let  the  polyhedra  be  placed  in  perspective  (§  259), 

0  the  center  of  similitude,  AyBxCx and  A2B2C2 

on  one  side  of   0,  and  A3B3C3 on  the  other. 

2.  Then  as  in  IV.  prop.  XX,  AXBX  II  A2B2  II  A3B3  and 
ZV^i  II  A4  II  I)ZA3 . 

3.  .'.  Z  B1AlI)1  =  Z  B2AJ)2  =  Z  B3A3D3 ,  and  similarly 
for  other  face  angles,  which  proves  (1).    VI,  prop.  V 

4.  The  trihedral  Z  Ax  =  Z  X  because  the  face  Z  are 
respectively  equal  and  similarly  placed,  and  is  sym- 
metric to  Z  A3  because  the  face  angles  are  respec- 
tively equal  and  placed  in  reverse  order. 

Prop.  XXI,  cor. 

5.  So  for  the  other  trihedral  Z.  And  V  polyhedral 
Z,  as  Dx,  D2,  D3,  can  be  cut  into  congruent  or 
symmetric  trihedral  A  similarly  placed,  as  by  the 
planes  AlC\D1,  A2C2D2,  A3C3D3,  they  too  are  con- 
gruent or  symmetric. 

6.  .'.  the  dihedral  Z  are  equal,  which  proves  (2),  and 
the  corresponding  polyhedral  Z  are  congruent  or 
symmetric,  which  proves  (3). 

7.  The  corresponding  edges,  as  AlBl,  A2B2,  A3B3, 
being  corresponding  sides  of  similar  A  OAxBx, 
OA2B2,  OA3B3,  have  the  ratio  of  similitude,  which 
proves  (4). 


366 


SOLID    GEOMETRY. 


[Bk.  VIII. 


Corollaries.     1.  If  the  ratio  of  similitude  is  1,  the  poly- 
hedra are  either  congruent  or  symmetric. 

2.    Corresponding  faces   of   similar  polyhedra  are  propor- 
tional to  the  squares  of  any  two  corresponding  edges. 
Step  7,  and  V,  prop.  IV. 


Proposition  XXX. 

501.  Theorem.  Tivo  similar  polyhedra  can  be  divided 
into  the  same  number  of  tetrahedra  similar  each  to  each 
and  similarly  placed. 

Proof.    1.  In  the  figure  below,  the  plane  of  Alf  Clf  Dl  and  the 
plane  of  A2,   C2,  D2  cut  off  tetrahedra  AlBiClD1, 

2.  Any  point  P1  in  the  one  has  a  corresponding  point 
P2  in  the  other,  such  that  OPx :  OP2  =  the  ratio  of 
similitude.  Why  ? 

3.  Hence  the  tetrahedra  are    similar. 


Proposition   XXXI. 

502.    Theorem.     The  volumes  of  similar  polyhedra  are  to 
each  other  as  the  cubes  of  their  corresponding  edges. 


Given         the    similar    polyhedra    AlBlC1  ,    A2I>.,(\,  , 

AtBsCg    having   volumes    vu  v.,,   vs,    respec- 
tively. 


Prop.  XXXI.]  SIMILAR    SOLIDS.  367 

To  prove    that  >\  :  r2  =  AXB?  :  J.,/;,'; .   >\  :  r3  =  .I^3 :  A%B£. 

Proof.    1.  Place  the  polyhedra  in  perspective,  as  in  the  figure, 
the  center  of  similitude  being  0. 
Divide  the  polyhedra  into  similar  tetrahedra,  simi- 
larly placed,  AXBXCXDX,  A2B2C2D2,  A3B3C3D3,  being 
corresponding  tetrahedra.  Prop.  XXX 

Let  tx,  t.2,  ts  represent  the  volumes  of  these  respec- 
tive tetrahedra,  pu  p2}  jh  their  altitudes  from  Dx, 
D2,  JJ3,  and  alf  a2,  a3  the  areas  of  AA1BlC1, 
A2B2C2,  A3B3C3. 

2.  Then     v*l  =  Jjpla1, 

and  t.2  =  i  2?2a2  > 

.\tl:t2   =     jh«i  -Ptfh' 

3.  But      a,  :  a,  =  AXBX2 :  A2B.22,  V,  prop.  IV 
and     px  : i>2  =  D1A1  :  D2A2  =  AXBX  :  A2B2. 

IV,  prop.  XX 

4.  .'.  pxax  '■  lha2  =  AXBXZ  :  A2B.? .  IV,  prop.  VII,  cor. 

5.  .'.  tv :  t2      =  AXBX3 :  A2B}.  From  steps  2,  4 

Similarly  the  other  tetrahedra  are  proportional  to 
the  cubes  of  their  corresponding  edges,  which  edges 
are  proportional  to  the  particular  edges  AXBX  and 
A2B2. 

6.  .'.  the  sum  of  the  tetrahedra  making  up  the  polyhe- 
dron A1B1C\ has  the  same  ratio  to  the  sum  of 

the  tetrahedra  making  up  the  polyhedron  A2B2C2 

as  AXBXZ  has  to  A2B.?,  or 

vt  :  v2  =  AXBX3 :  A2B3. 

Similarly  vx  :  r3  =  AXBX3 :  A3B33, 

v2  '.  v3  =  A2£>2  '•  A3u3 , 

=  Jj2i^2   '.  Jj3C3  , 

=  C2D3 :  C3D3,  etc. 


368 


SOLID    GEOMETRY. 


[Bk.  VIIL 


Proposition  XXXII. 

503.    Theorem.     Any  two  spheres  are  similar. 

Proof.        Let  the  spheres  be  placed  in  concentric  position. 

Then  V  the  ratio  of    their  radii    is    constant  any 
point  on  the  surface  of  the  one  has  on  the  surface 
of  the  other  its  similar  point,  with  respect  to  the 
center,  the  ratio  being  i\  :  r2. 
.'.  the  spheres  are  similar. 


Proposition  XXXIII. 

504.    Theorem.     Two  right  circular  cylinders  are  similar 
if  their  elements  have  the  same  ratio  as  the  radii  of  the ir 

bases. 


Proof.  1.  Let  the  cylinders  have  the  radii  r19  r.2,  and  the  alti- 
tudes hlt  h2,  respectively,  and  be  placed  with  their 
axes  in  the  same  line,  their  mid-points  coincid- 
ing at  0. 

Let  the  semi-altitudes  be  OA1}  OJ.2,  and  let  aline 
from  0  cut  the  bases  in  Blf  B2,  not  necessarily 
on  the  circumferences,  and  one  from  0  cut  the 
cylindrical  surf  aces  in    Clt   C2,  respectively. 

2.   Then  '•'  the  altitudes  are  proportional  to  the  radii, 


Prop.  XXXIV.]  SIMILAR   SOLIDS.  369 

.'.  OAj  :  OA2=  rx:r2. 
And  ••' AXBX\\A.2B,,  Why? 

.-.OB,:  OB,,  =  OA1  :  OA,  =  r,  :  r2.  IV,  prop.  X 

3.  And  '•'  the  axes  coincide  .\  the  elements  are  parallel, 
and  .*.  OC,:  OC,  =  rx:r2. 

.'.  the  points  of  the  respective  cylinders  are  similar 
with  respect  to  0  as  a  center. 

Corollaries.  1.  The  areas  of  the  cylindrical  surfaces  of 
two  similar  cylinders  are  proportional  to  the  squares  of  their 
altitudes. 

For  «i  =  2  7tri7ii  and  a2  =  2  7tr2h2. 

ai  _  rdii 
'  '  a2      foho 

But  v  ^  =  ^  by  prop.  XXXIII, 

' '  a2      h22 

2.  The  volumes  of  two  similar  rigid  circular  cylinders  are 
proportional  to  the  cubes  of  their  altitudes. 


Proposition  XXXIV. 

505.    Theorem.     Two  right    circular  cones    are  similar    if 

their  altitudes   have  the   same    ratio  as  the  radii  of   their 

bases. 

Place  the  bases  in  concentric  position.  The  proof  is  then  so  similar  to 
that  of  prop.  XXXIII  that  it  is  left  for  the  student. 

Corollaries.  1.  The  areas  of  the  surfaces  of  two  similar 
right  circular  cones  are  proportional  to  the  squares  of  their 
altitudes. 

2.  The  volumes  of  two  similar  right  circular  cones  are  pro- 
portional to  the  cubes  of  their  altitudes. 


370  SOLID   GEOMETRY.  [Bk.  VIII.] 


EXERCISES. 

750.  The  mean  radii  of  the  earth  and  moon  are  respectively  3956 
miles,  1080.3  miles.     Show  that  their  volumes  are  as  49  to  1,  nearly. 

751.  The  mean  diameter  of  the  planet  Jupiter  being  80,657  miles, 
find  the  ratio  of  its  volume  to  that  of  the  earth. 

752.  The  sun's  diameter  is  about  109  times  the  earth's.  Find  the 
ratio  of  their  volumes. 

753.  What  is  the  radius  of  that  sphere  whose  number  of  square  units 
of  surface  equals  the  number  of  cubic  units  of  volume  ? 

754.  Also  of  that  whose  number  of  cubic  units  of  volume  equals  the 
number  of  square  units  of  area  of  one  of  its  great  circles. 

755.  Also  of  that  whose  number  of  cubic  units  of  volume  equals  the 
number  of  linear  units  of  the  circumference  of  a  great  circle. 

756.  Two  planes  cut  a  sphere  of  radius  1  m,  at  distances  0.8  m  and 
0.5  m  from  the  center.  Find  (1)  the  area  of  the  zone  between  them, 
(2)  the  volume  of  the  corresponding  spherical  segment. 

757.  A  solid  cylinder  20  cm  long  and  2  cm  in  diameter  is  terminated 
by  two  hemispheres.  The  solid  is  melted  and  molded  into  a  sphere. 
Find  the  diameter  of  the  sphere. 

758.  A  meter  was  originally  intended  to  be  0.0000001  of  a  quadrant 
of  the  circumference  of  the  earth.  Assuming  it  to  be  such,  and  the  earth 
to  be  a  sphere,  find  its  radius  in  kilometers. 

759.  A  cone,  a  sphere,  and  a  cylinder  have  the  same  altitudes  and 
diameters.     Show  that  their  volumes  are  in  arithmetical  progression. 

760.  Given  a  sphere  of  radius  10.  How  far  from  its  center  must  the 
eye  be  in  order  to  see  one-fourth  of  its  surface  ? 

761.  If  a  tetrahedron  is  cut  by  a  plane  parallel  to  one  of  its  faces,  the 
tetrahedron  cut  off  is  similar  to  the  first. 

762.  The  areas  of  the  surfaces  of  two  similar  polyhedra  are  proportional 
to  the  squares  of  their  corresponding  edges. 


NUMERICAL    TABLES.  371 


506.     NUMERICAL   TABLES. 

Formula  or  Mensuration.  The  numbers  refer  to  the  pages.  Ab- 
breviations :  b,  base  ;  h,  altitude  ;  r,  radius  ;  a,  area ;  c,  circumference  ; 
p.  perimeter;  s,  slant  height;  »,  volume;  m,  mid-section. 

Parallelogram,  202,  a  =  bh.  Circle,  217,  224,  c  =  27rr. 

Triangle,  202,  a  =  \  bh.  a  =  nr2. 

Trapezoid,  202,  a  =  $  (b  +  b')  h.  Arc,  223,     =  a  •  r. 

Parallelepiped,  307,  v  =  bh. 

Prism,  307,  v  —  bh. 

Lateral  area,  right  prism,  298,  a  =  ph. 
Prismatoid,  314,  v  =  \h{b  +  6'  +  4 m). 

Pyramid,  313,  v  =  %bh. 

Lateral  area,  regular  pyramid,  309,  a  =  ^s. 
Frustum  of  pyramid,  315,  v  =  i-  h  (b  +  b'  +  V&&'). 

Lateral  area,  frustum  of  regular  pyramid,  309,  a  =  £  (p  +  p')  s. 
Right  circular  cylinder,  324,  325,  v  =  bh  =  xr-h.     Lateral  a  =  ch  =  2  nrh. 
Right  circular  cone,  324,  325,        v  =  \bh  =  J  7rr2/i. 

Lateral  a  =  £  cs  =  7frs. 
Frust.  of  rt.  circ.  cone,  325,  v  —  ^ith  (i\2  +  r22  +  rir2). 

Sphere,  355,  360,  v  =  1 7rr3.         a  =  4  tzT2. 

Lune,  356,  a  =  2  ar2. 

Spherical  polygon,  356,  a  =  ar2. 

Zone,  356,  a  =  2  ;rWL 

Spherical  segment,  363,  v  =  \nh [3 (n«  +  r22)  +  ft2]. 

Spherical  sector,  362,  v  =  f  7Tr2ft  =  £&r. 

Most  Important  Expressions  involving  7T. 

*  =  3.141693.  1/7T  =  0.31830989.  180°/ ?r  =  57°.29578. 

7T/4  =  0.785398.  tt2  =  9.86960440.  tt/180  =     0.01745. 

?r/3  =  1.047198.  V#  =  1.77245385.  Approximate  values  ; 

|  tt  =  4.188790.  1/Var  =  0.56418958. 

Certain  Numerical  Results  frequently  used. 

V2  =  1.4142.  VlO  =  3.1623.  VE  =  1.7100. 

V3=  1.7821.  VJ    =0.7071.  V6  =1.8171. 

V5  =2.2361.  V2    =1.2599.  V7  =1.9129. 

V6  =  2.4495.  V3    =  1.4422.  V9  =  2.0801. 


3  " 

V7  =2.6458.  VI    =1.5874.  VlO 


2.1544. 


372  BIOGRAPHICAL    TABLE. 


507.     BIOGRAPHICAL   TABLE. 

The  following  table  includes  only  those  names  mentioned  in  this  work, 
although  numerous  others  might  profitably  be  considered  by  the  student. 
The  history  of  geometry  may  be  said  to  begin  in  Egypt,  the  work  of 
Ahmes,  copied  from  a  treatise  of  about  2500  B.C.,  containing  numerous 
geometric  formulae.  The  scientific  study  of  the  subject  did  not  begin, 
however,  until  Thales  visited  that  country,  and  carried  the  learning  of 
the  Egyptians  back  to  Greece.  The  period  of  about  four  hundred  years 
from  Thales  to  Archimedes  may  be  called  the  golden  age  of  geometry. 
The  contributions  of  the  latter  to  the  mensuration  of  the  circle  and  of 
certain  solids  practically  closed  the  scientific  study  of  the  subject  in 
ancient  times.  Only  a  few  contributors,  such  as  Hero,  Ptolemy,  and 
Menelaus,  added  anything  of  importance  during  the  eighteen  hundred 
years  which  preceded  the  opening  of  the  seventeenth  century.  Within 
the  past  three  hundred  years  several  important  propositions  and  numerous 
improvements  in  method  have  been  added,  but  the  great  body  of  ele- 
mentary plane  geometry  is  quite  as  Euclid  left  it.  In  recent  times  a 
new  department  has  been  created,  known  as  Modern  Geometry,  involving 
an  extensive  study  of  loci,  collinearity,  concurrence,  and  other  subjects 
beyond  the  present  range  of  the  student's  knowledge. 

The  pronunciations  here  given  are  those  of  the  Century  Cyclopedia  of 
Names.  The  first  date  indicates  the  year  of  birth,  the  second  the  year  of 
death.  All  dates  are  a.d.  unless  the  contrary  is  indicated  by  the  sign  — . 
The  letter  c.  stands  for  circa,  about,  b.  for  born,  d.  for  died.  Numbers 
after  the  biographical  note  refer  to  pages  in  this  work. 

Key.     L.  Latin,  G.  Greek,  dim.  diminutive,  fern,  feminine, 
a  fat,        a  fate,        a  far,        a  fall,        a  ask,        a  fare, 
e  met,        e  mete,        e  her,         i  pin,         i  pine,       o  )i<>/. 
o  note,       6  move,       6  nor,        u  tub,         u  mute,      u  pull. 
h  French  nasalizing  n.  ch  German  ch. 

s  as  in  leisure.  t  as  in  nature. 

A  single  dot  under  a  vowel  indicates  its  abbreviation. 
A  double  dot  under  a  vowel  indicates  that  the  vowel  approaches  the 
short  sound  of  u,  as  in  put. 


BIOGRAPHICAL    TABLE.  373 

Ahmes  (a'mes).     c.  —  1700.      Egyptian  priest      Wrote   the   oldest 

extant  work  on  mathematics 221 

Anaxagoras  (an-aks-ag'0-ras).      -  499,    -  -428.     Greek  philosopher 

and  mathematician 225 

Archimedes  (ar-M-mS'dk).     -287,    -212.    Syracuse,  Sicily.    The 

greatest  mathematician,  and  physicist  of  antiquity  .  87.  221,  353,  354 
Aryabhatta    (ar-ya-bha'ta).      b.   470.     One    of   the   earliest  Hindu 

mathematicians.     Wrote  on  Algebra  and  Geometry 221 

Bhaskara  (bhas'ka-ra).      12th  cent.     Hindu  mathematician      .     .     .   104 
Brahmagupta    (brah-ma-gop'ta).      b.    508.      Hindu   mathematician. 
One  of  the  earliest  Indian  writers 143,  221 

Carnot  (kar-no'),  Lazare  Nicholas  Marguerite.  1753,  1823.  French 
physicist  and  mathematician.  Contributed  to  Modern  Geom- 
etry          241.  242 

Oavalieri  (ka-va-le-a'iv).  Bonaventnra.  1598,  1647.  Prominent  Ital- 
ian mathematician 351 

Ceulen  (koilen).  Ludolph  van.     1540.  1010.     Dutch  geometrician    .  221 
Ceva  (cha'va),  Giovanni.     1048,  c.   1737.     Italian  geometrician,  239.  241 
Dase  (da'ze),  Zacharias.     1824,  1861.     Famous  German  computer    .  221 
Descartes  (da-kart'),  Kene.     1596,   1650.     Eminent  French  mathe- 
matician, physicist,  and  philosopher.     Founder  of  the  science  of 

Analytic  Geometry 285 

Euclid  (u'klid).  c.  —  300.  Eminent  writer  on  Geometry  in  the 
Alexandrian  School,  at  Alexandria,  Egypt,  His  "Elements," 
the  first  scientific  text-book  on  the  subject,  is  still  the  standard 

in  the  schools  of  England 76,  152,  208 

Euler  (oiler),  Leonhard.     1707,  1783.      Swiss.     One  of  the  greatest 

mathematicians  of  modern  times        99,  108,  285,  289 

Gauss  (gous),  Karl  Friedrich.     1777.    1855.     German.    One  of  the 

greatest  mathematicians  of  modern  times 208,  212 

Hero  (he'ro)of  Alexandria.    More  properly  Heron  (he'ron).    c.  —  110. 

Celebrated  Greek  surveyor  and  mechanician        221.  227 

Hippocrates  (hi-pok'ra-tez)  of  Chios,     b.  c.  —  470.     Author  of  the 

first  elementary  text-book  on  Geometry 230 

Jones  (jonz),  William.     1675-1749.     English  teacher 221 

Klein  (kiln),  Christian  Felix.     1849.     Professor  at  Gottingen       .     .  225 
Leibnitz  (lib'nits),  Gottfried  Wilhelm.     1646,  1716.     Equally  cele- 
brated as  a  philosopher  and  a  mathematician.    One  of  the  founders 

of  the  science  of  the  Calculus 23,  182 

Lindemann  (lin'de-man),  Ferdinand,      b.   1852.     German  professor  225 


374  BIOGRAPHICAL    TABLE. 

Meister  (mis'ter),  Albrecht.     1724-1788.     German  mathematician    .     98 

Menelaus  (men-e-la/us).  c.  100.  Greek  mathematician  and  astrono- 
mer.    One  of  the  early  writers  on  Trigonometry     .     .      240,  242,  243 

Metius  (metius).  Anthonisz,  Adrisen.  Called  Metius  from  Metz, 
his  birthplace.     1527-1607 221 

Monge  (mohzh),  Gaspard.  1746,  1818.  French.  Founder  of  the 
science  of  Descriptive  Geometry.  One  of  the  founders  of  the 
Polytechnic  School  of  Paris 97 

CBnopides  (e-nop'i-dez).     c.  -  465.     Early  Greek  Geometer    ...     72 
Pascal  (pas-kal'),  Blaise.    1623,  1662.    Celebrated  French  mathemati- 
cian, physicist,  and  philosopher 241 

Plato  (pla'to).     c.  —  429,  —  348.     Greek  philosopher  and  founder 

of  a  school  that  contributed  extensively  to  Geometry,  68,  106,  152,  286 
Pothenot  (po-te-no'),  Laurent,     d.  1732.     French  professor     .     .     .157 
Ptolemy    (tore-mi).      Claudius  Ptolemseus.      87,   165.     One  of   the 
greatest  of  astronomers,  geographers,  and  geometers  of  the  later 

Greeks 221,  228 

Pythagoras  (pi-thag'o-ras).  c.  —  580,  c.  —  501.  Founder  of  a  cele- 
brated school  in  Lower  Italy.  One  of  the  foremost  of  the  early 
mathematicians 49,  103,  286 

Richter (rich'ter).     1854.     German  computer 221 

Thales  (tha'lez).  -  640,  -  548.  One  of  the  Seven  Wise  Men  of 
Greece.     Introduced  the  study  of  Geometry  from  Egypt,   26,  117,  131 

Vega,  Georg,  Freiherr  von.  1756-1802.  Professor  of  mathe- 
matics at  Vienna 221 


TABLE   OF   ETYMOLOGIES. 


This  table  includes  such  of  the  pronunciations  and  etymologies  of  the 
more  common  terms  of  Geometry  as  are  of  greatest  value  to  the  student. 
The  equivalent  foreign  word  is  not  always  given,  but  rather  the  primitive 
root  as  being  more  helpful.  The  pronunciations  and  etymologies  are 
those  of  the  Century  Dictionary.     See  Biographical  Table,  p.  372. 


Abscissa  (ab-sis'a).     L.  cut  off. 
Acute  (a-kut').     L.  acutus,  sharp. 
Adjacent  (a-ja'sent).    L.  ad,  to,  + 

jacere,  lie. 
Angle  (ang'gl).     L.  angulus,  a  cor- 
ner, an  angle  ;  G.  ankylos,  bent. 
Antecedent  (an-te-se'dent).  L.  ante, 

before,  +  ceclere,  go. 
Bisect  (bi-sekf).     L.  6*'-,  two-,   + 

sectus,  cut. 
Center  (sender).  L.  centrum,  center; 

G.  kentron,  from  kentein,  to  prick. 
Centroid  (sen'troid).     G.    kentron, 

center,  +  eiclos,  form. 
Chord  (kord).     G.  chorde,  string. 
Circle  (sir'kl).    L.  circulus,  dim.  of 

circus  (G.  kirkos),  a  ring. 
Circumference     (ser-kum'f  e-rens) . 

L.   circum,  around  (see  Circle), 

+  ferre,  to  bear. 
Collinear     (ko-lin'e-ar).     L.  com-, 

together,  +  linea,  line. 
Commensurable  (ko-men'su-ra-bl) , 

L.  com-,  together,  +  mensurare, 

measure. 
Complement   (kom'ple-ment).      L. 

complementum,  that  which  fills, 

from  com-  (intensive)  +  plere,  fill. 


Concave  (kon'-kav).  L.  com-  (in- 
tensive) +  cavus,  hollow. 

Concentric  (kon-sen'trik).  L.  con-, 
together,  +  centrum,  center. 

Concurrent  (kon-kur'ent).  L.  con-, 
together,  +  currere,  run. 

Concyclic  (kon-sik'lik).  L.  con-, 
together,  +  cyclicus,  from  G. 
kyklikos,  from  kyklos,  a  circle, 
related  to  kyliein,  roll  (compare 
Cylinder). 

Congruent  (kong/gro-ent).  L.  con- 
gruere,  to  agree. 

Consequent  (kon'se-kwent).  L. 
con-,  together,  +  sequi,  follow. 

Constant  (kon'stant).  L.  con-,  to- 
gether, +  stare,  stand. 

Converse  (kon'vers).  L.  con-,  to- 
gether, +  vertere,  turn. 

Convex  (kon'veks).  L.  convexus, 
vaulted,  from  con-,  together,  -f 
vehere,  carry. 

Corollary  (kor'o-la-ri).  L.  corolla- 
Hum,  a  gift,  money  paid  for  a 
garland  of  flowers,  from  corolla, 
dim.  of  corona,  a  crown. 

Cylinder  (sirin-der).  G.  kyllndros, 
from  kyliein,  roll. 


375 


376 


TABLE    OF  ETYMOLOGIES. 


Decagon  (dek'a-gon).    G.  deka,  ten, 

+  gonia,  an  angle. 
Degree  (de-gre').     L.  de,  down,  4- 

gradus,  step. 
Diagonal     (dl-ag'o-nal).      G.    dia, 

through,  +  gonia,  a  corner,  an 

angle. 
Diameter    (di-am'e-ter).      G.    dia, 

through,  4-  metron,  a  measure. 
Dihedral  (dl-he'dral).    G.  di-,  two, 

+  hedra,  a  seat. 
Dimension  (di-men'shpn).    L.  dis-, 

apart,   4-  metiri,  measure.     See 

Measure. 
Directrix    (di-rek'triks).     L.    fem. 

of  director,  from  directus,  direct. 
Dodecahedron  (do"dek-a-he'drpn). 

G.  dodeka,  twelve,   +  hedra,  a 

seat. 
Equal  (e'kwal).    L.  cequalis,  equal, 

from  cequus,  plain. 
Equiangular  (e-kwi-ang'gu-liir).  L. 

cequus,  equal,  +  angulus,  angle. 
Equilateral     (e-kwi-lat'e-ral).      L. 

cequus,  equal,  +  latus,  side. 
Equivalent      (e-kwiv'a-lent).       L. 

cequus,  equal,  4-  valere,  be  strong. 
Escribed  (es-krlbd')-     L.  e,  out,  + 

scribere,  write. 
Excess   (ek-ses').     L.   ex,   out,    -f 

cedere,  go  ;  i.e.  to  pass  beyond. 
Frustum  (frus'turn).  L.  a  piece. 
Generatrix  (jen'e-ra-triks).  L.  fem. 

of  generator,  from  generare,  beget, 

from  genus,  a  race. 
Geometry  (je-om'e-tri).     G.  ge,  the 

earth,  +  metron,  a  measure, 
-gon,  a  termination,  G.  gonia,  an 

angle. 
Harmonic   (har-mon'ik).     G.  har- 

monia,  a  concord,  related  to  har- 

mos,  a  joining.     A  line  divided 


internally  and  externally  in  the 
ratio  2:1,  is  cut  into  segments 
representing  1,  f,  £.  Pythago- 
ras first  discovered  that  a  vibrat- 
ing string  stopped  at  half  its 
length  gave  the  octave  of  the 
original  note,  and  stopped  at 
two-thirds  of  its  length  gave 
the  fifth.  Hence  the  expression 
"harmonic  division  "  of  a  line. 

Hemisphere  (hem'i-sfer).  G.  hemi-, 
half,  +  sphaira,  a  sphere. 

-hedron,  a  termination,  G.  hedra, 
a  seat. 

Hepta-,  in  combination,  G.  seven. 

Hexa-,  in  combination,  G.  six. 

Hexagram  (hek'sa-gram).  G.  hex, 
six,  4-  gramma,  a  line. 

Hypotenuse  (hi-pot'e-nus).  G.  hypo, 
under,  4-  teinein,  stretch. 

Inclination  (in-kli-na'shon).  L.  in, 
on,  4-  clinare,  lean. 

Incommensurable  (in-ko-men'su- 
ra-bl).  L.  in-,  not,  4-  com-,  to- 
gether, 4-  mensurare,  measure. 

Infinity  (in-fin'i.-ti).  L.  in-,  not, 
4  finitus,  bounded. 

Inscribed  (in-skribd7).  L.  in,  in, 
4-  scribere,  write. 

Isosceles  (i-sos'e-lez).  G.  isos, 
equal,  -f  skelos,  leg. 

Lateral  (lat'e-ral).  L.  latus,  a 
side. 

Locus  (lo'kus).  L.  a  place.  Com- 
pare locality. 

Lune  (lun).     L.  luna,  the  moon. 

Major  (ma/jor).  L.  greater,  com- 
parative of  magnus,  great. 

Maximum  (mak'si-mum).  L.  great- 
est, superlative  of  magnus, 
great. 

Mean  (men).     L.  medius,  middle. 


TABLE    OF  ETYMOLOGIES. 


;;:: 


Measure  (mezh'iir).      L.  mensura, 

a  measuring.     See  Dimension. 
Median  (me'di-au).     See  Mean. 
Mensuration  (men-su-ra'shon).    See 

Measure. 
Minimum  (min'i-nium).     L.  least. 
Minor  (mi 'nor).     L.  less. 
Nappe    (nap).      French,    a    cloth, 

sheet,  surface. 
Oblique  (ob-lek7  or  ob-Hk').    L.  ob, 

before,  +  liquis,  slanting. 
Obtuse  (ob-tus').    L.  obtusus,  blunt, 

from  ob,  upon,  +  twidere,  strike. 
Octo-,  octa-,  in  combination,  L.  and 

G.,  eight. 
Opposite  (op'o-zit).     L.  ob,  before, 

against,  +  ponere,  put,  set. 
Ordinate  (or'di-nat).     L.  ordo  (or- 

din-),  a  row. 
Orthocenter    (Or'tho-sen-ter).       G. 

ortho-,  straight,  -j-  kentron, center. 
Orthogonal     (or-thog'o-nal).        G. 

orthos,  right,  +  gonia,  an  angle. 
Parallel  (par'a-lel).     G.  para,  be- 
side, +  allelon,  one  another. 
Parallelepiped  (par-a-lel-e-pip'ed  or 

-pl'ped).     Gr.  parallelos,  parallel, 

+  epipedon,  a  plane  surface,  from 

epi,  on,  -f  pedon,  ground. 
Parallelogram      (par-a-lel'S-grani) . 

G.  parallelos,  parallel,  +  gram  ma, 

line. 
Pedal  (ped'al  or  pe'dal).     L.  peda- 

lis.  pertaining  to  the  foot,  from 

pes  (ped-).  foot. 
Pencil  (pen'sil).      L.  penic  ilium,  a 

painters1  pencil,  a  brush. 
Perigon     (per'i-gon).         G.     peri, 

around,  +  gonia,  a  corner,  angle. 
Perimeter  (pe-rim'e-ter).     G.  peri, 

around,  +  metron,  measure. 
Perpendicular    (per-pen-dik  u-lar). 


L.  perpendiculum ,  a  plumb-line, 
from  per,  through,  -f  pcndere. 
hang. 

Perspective  (per-spek'tiv).  L.  per, 
through,  +  specere,  see. 

it  (pi).  Initial  of  G.  periphereia, 
periphery,  circumference. 

Pole  (pol).  G.  polos,  a  pivot,  hinge, 
axis,  pole. 

Polygon  (pol'i-gon).  G.  polys, 
many,  +  gonia,  corner,  angle. 

Polyhedron,  (pol-i-he'drpn)  G. 
polys,  many,  +  hedra,  seat. 

Postulate  (pos'tu-lat).  L.  postula- 
tum,  a  demand,  from  poscere, 
ask. 

Prism  (prizm).  G.  prisma,  some- 
thing sawed,  from  priein.  saw. 

Prismatoid  (priz'ma-toid).  G.  pris- 
ma (t~),  +  eidos,  form. 

Projection  (pro-jek'skpn).  L.  pro, 
forth,  +  jacere,  throw. 

Pyramid  (pir'a-mid).  G.  pyramis, 
a  pyramid,  perhaps  from  Egyp- 
tian pir-em-us,  the  slanting  edge 
of  a  pyramid. 

Quadrant  (kwod'rant).  L.  quad- 
rants, a  fourth  part.  See 
Quadrilateral. 

Quadrilateral  (kwod-ri-lat'e-ral).  L. 
quattuor  (quadri-),  four,  -f  latus, 
(later-),  side. 

Radius  (ra'di-us).  L.  rod,  spoke 
of  a  wheel. 

Ratio  (ra'shio).  L.  a  reckoning, 
calculation,  from  reri,  think, 
estimate. 

Reciprocal  (re-sip'ro-kal).  L.  re- 
ciprocus.  returning,  from  re-, 
back,  and  pro.  forward,  with 
two  adjective  terminations. 

Rectangle  (rek'tang-gl).    L.  rectus, 


378 


TABLE    OF  ETYMOLOGIES. 


right,  +  angulus,  an  angle.  See 
Angle. 

Rectilinear  (rek-ti-lin'-ar).  L.  rec- 
tus, right,  +  linea,  a  line. 

Reflex  (re'fleks  or  re-fleks').  L. 
re-,  back,  -f  flectere,  bend. 

Regular  (reg/u-lar).  L.  regula,  a 
rule,  from  regere,  rule,  govern.. 

Rhombus  (rom'bus).  G.  rhombos, 
a  spinning  top. 

Scalene  (ska-len').  G.  skalenos, 
uneven,  unequal ;  related  to  skel- 
los,  crooked-legged. 

Secant  (se'kant).  L.  secare,  cut, 
as  also  Sector,  Section,  Seg- 
ment. 

Segment  (seg'ment).     See  Secant. 

Semicircle  (sem'i-ser-kl).  L.  semi-, 
half,  +  circulus,  circle. 

Similar  (sim'i-lar).    L.  similis,  like. 

Solid  (sol'id).  L.  solidus,  firm, 
compact. 

Sphere  (sfer).     G.  sphaira,  a  ball. 

Square  (skwar).  L.  quadra,  a 
square,  from  quattuor,  four. 

Straight  (strat).  Anglo-Saxon, 
streht,  from  streccan,  stretch. 

Subtend  (sub-tend').  L.  sub,  under, 
-f  tender e,  stretch. 

Successive  (suk-ses'iv).  L.  sub,  un- 
der, +  cedere,  go. 

Sum  (sum).  L.  summa,  highest 
part.     Compare  Summit. 

Superposition  (su'per-po-zish'qn). 
L.  super,  over,  +  ponere,  lay. 


Supplement  (sup'le-ment).  L.  sub., 
under,  +  plere,  fill ;  to  fill  up. 

Surface  (ser-fas).  L.  superficies, 
surface,  from  super,  above,  + 
fades,  form,  figure,  face. 

Symbol  (sim'bpl).  G.  symbolos,  a 
sign  by  which  one  infers  some- 
thing, from  sun,  together,  +  bal- 
lein,  put. 

Tangent  (tan'j.ent).  L.  tangere, 
touch. 

Tetrahedron  (tet-ra-he'drpn).  G. 
tetra-,  four,  -|-  hedra,  seat. 

Theorem  (the'o-rem).  G.  theorema, 
a  sight,  a  principle  contemplated. 

Transversal  (trans-ver'sal).  L. 
trans,  across,  -f  vertere,  turn. 

Trapezium  (tra-pe'zi-um).  G.  tra- 
pezion,  a  table,  dim.  of  trapeza, 
a  table,  from  tetra,  four,  +  pous, 
foot. 

Trapezoid  (tra-pe'zoid).  G.  trapeza, 
table,  +  eidos,  form. 

Tri-,  in  composition,  L.  tres  (tri-), 
G.  treis  (tri-),  three.  See  Secant, 
-hedron,  Angle,  for  meaning  of 
trisect,  trihedral,   triangle. 

Truncate  (trung'kat).  L.  truncare, 
cut  off,  from  Old  L.  troncus,  cut 
off,  mutilated. 

Unique  (u-nek').  L.  unicus,  from 
unus,  one. 

Vertex  (ver'teks).    L.  vertere,  turn. 

Zone  (zon).  G.  zone,  a  girdle, 
belt. 


INDEX. 


PAGE 

Abbreviations 12 

Acute  angle 6,  266 

Acute-angled  triangle     ...  50 

Adjacent  angles    ....  6,  266 

"       polygons     ....  90 

Alternate  angles 43 

Alternation 163 

Altitude  90,  295,  313,  321,  350,  361 

Ambiguous  case 51 

Analysis 70,  152 

Angle  .    5,  115,  151,  256,  261, 

265,  334,  335 

Antecedent 161 

Antiparallel 177 

Apothem 214 

Arc 67,  114 

Area 112,  202 

Arm 5 

Assumed  constructions  ...  70 

Axioms 11 

Axis   ....    29,  265,  321,  328 
Base,  23,  29,  59,  292,  295,  321, 

350,  361 

Birectangular  triangle    .     .     .  345 

Bisect 4,  5,  266 

Broken  line -  .  4 

Cavalieri's  theorem  ....  359 

Center 67,  114,  326 

"      line 143 

"      of  mass  (gravity)      .     .  89 

"      of  polygon       ....  214 

"      of  similitude   ....  183 


PAGF 

Center  of  symmetry       .     .     .183 

"      segment 143 

Central  angle .115 

"       symmetry      .     .     .     .183 

Centroid 89 

Chord 114 

Circle 67,  114.  327 

Circular  cone   .     .     .     .321,  322 

"       cylinder 318 

Circumcenter 89 

Circumference  ...  67,  114 
Circumscribed,  136,  212,  322,  334 

Collinear 84,  238 

Commensurable 160 

Complement     .     .     .    8,  115,  266 

Composition 164 

Concave 54,  283,  335 

Concentric 143 

Concurrent 84,  238 

Coney  clic 128 

Cone 321,  322,  361 

"    of  revolution     ....  322 

Congruent 23,  DO 

Conical  surface,  space  .  .  .321 
Conjugate  .  .  .  .  8,  115,  266 
Consecutive  angles    ....     59 

Consequent 161 

Constant 167 

Contact 125,  332 

Converse 34 

Convex  .  .  .  .  7,  54,  283,  335 
Coplanar 245 


379 


380 


INDEX. 


Corollary 20 

Corresponding  angles     ...     43 
"         segments,  points, 

179,  182 
Counter-clockwise 
Cross  polygons 
Curve      .... 
Curved  surface 
Cylinder      .     .     . 

"       of  revolution     . 
Cylindrical  surface,  space 
Degree    .     . 


Determined 
Diagonal 
Diagonal  scale 
Diameter  . 
Difference  . 
Dihedral 
Dimensions 
Directrix 
Distance 
Distributive  law 
Division  .  .  . 
Division,  internal 


o 

.     54 

.     67 

.  317 

.  318 

.  322 

.  317 

5,  222 

.  3,  25,  51 

.      22,  288 

.     .     .184 

67,  114,  326 

90,   115,  266 

.     .     .  265 

.     .     .  305 

.    317,  321 

29,  259,  264 

...     96 

.     164,  170 

and  external, 

95,  176 
Edges      .  265,  274,  283,  291,  294 

Element 317,  321 

Equal      .     .      6,  24,  90,  114,  222 

Equiangular 22 

Equilateral 22 

Equivalent 24,  90 

Escribed      .......  136 

Ex-center 89 

Excess,  Spherical  .  .  .  .353 
Exterior  angles  .  .  .  .  22,  43 
Extreme  and  mean  ratio     .     .   196 

Extremes 161 

Faces      .  265,  274,  283,  291,  294 

Figure 21 

Fourth  proportional  ....  166 
Frustum 295,  322 


PAGE 

General  polygon 54 

Generalization  of  figures    .     .     56 

Generatrix 317,  321 

Geometry 1,  9,  21 

Geometric  mean 166 

Golden  section      .     .     .    196,  197 

Great  circle 327 

Harmonic  division     ....   176 

Hemisphere 328 

Hero's  formula 227 

Hexagram,  Mystic     ....  241 

Hypotenuse 50 

In-center 89 

Inclination 256 

Incommensurable       .     .     .     .160 

Indirect  proof 14 

Infinity 170,  174 

Inscribed,  128,  136,  157,  212, 

322,  334 
Instruments  of  geometry  .  .  208 
Interior  angles      .     .     .     .  22,  43 

Inversion 163 

Isoperimetric 229 

Isosceles 28,  59 

Lateral  area 298 

Law  of  Converse 34 

Limit       ....      167,  216,  323 

Line 2,  3 

Locus 80,  156 

Ludolphian  number  .     :     .     .221 

Lune 335 

Major 115,  128 

Maximum 229 

Mean  proportional     ....  166 

Means 161 

Measure,    112,  117,  129,    159, 

202,  266,  356 

Median 28 

"      section 196 

Mensuration,     112,  226,  298,  356 
Methods  of  attack    .    35,70,152 


INDEX. 


381 


PAGE 

Mid-point 4 

Minimum 229 

Minor 115,  128 

Motion    ........     24 

Mutually  equiangular,  etc.  .  22 
Mystic  hexagram       ....  241 

Nappes 321 

Negative  ....  56,  97,  98 
Nine  points  circle  ....  243 
Number,    101,  159,  166,  201,  306 

Oblique 7,  252,  266 

"       circular  cone      .     .     .  321 

"      cylinder 318 

"       prism 291 

Obtuse  angle 7,  266 

Obtuse-angled  triangle  ...  50 
One-to-one       correspondence, 

101,  159 
Opposite       ...  8,  59,  275,  339 

Orthocenter 89 

Orthogonal 104 

Parallel  .     .     .      43,  46,  262,  270 

Parallelepiped 288 

Parallelogram 59 

Parts  of  polygon 22 

Pedal  triangle 137 

Pencil      ....      170,  265,  270 

Perigon 5 

Perimeter 21 

Perpendicular  .       7,  252,  266,  352 

Perspective 183 

tc 217,  220-222 

Plane 4,  244 

"     figure,  Geometry  ...     21 
"     of  symmetry    ....  289 

"     section 326 

Platonic  bodies 286 

Point 2 

Polar  distance 329 

"     triangle,  polyhedral 

angle    .     .     .     .342,  345 


PAGE 

Pole 328 

Polygon  .     .     .     21,  54,  335,  339 
Polyhedral  angle  .     .     .    271,  339 

Polyhedron 283 

Positive  and  negative     .     .  56,  97 
Postulates,    9,  10,  24,  46,  68, 

216,  245,  318,  322,  326 
Preliminary  propositions     .     .     13 

Prism 292 

Prismatic  surface,  space     .     .  291 

Prismatoid 313 

Problem 9 

Produced 4 

Product  of  lines    .      166,  201,  306 
Projection    .     .     .      104,  255,  256 

Proof 19,  35 

Proportion 161 

Proposition 9 

Pyramid 295 

Pyramidal  surface,  space   .     .  294 
Pythagorean    Theorem,    102, 

103,  211 

Quadrant 115,  329 

Quadrature  of  the  circle     .     .  225 

Quadrilateral 22 

Radian    ........  222 

Radius    ...    67,  114,  214,  326 

Ratio 159,  184 

Reciprocal  Theorems,  27,  101,  340 

Rectangle 60,  94 

Rectangular  parallelepiped      .  289 

Rectilinear  figure 21 

Reductio  ad  absurdum  .      14,  152 

Reflex  angle 7 

Regular  polygon 54 

"       polyhedron    .     .    283,  286 

"      pyramid 308 

Rhombus 60 

Right  angle 7,  266 

Right-angled  triangle     ...     50 
Right  circular  cone    ....  321 


382 


INDEX. 


PAGE 

Right  section    ....    291,  317 

Rotation 24 

Scalene 29 

Secant 114 

Sector 115,  361 

Segment       ...  3,  95,  128.  356 

Semicircle 117 

Semicircumference    ....  115 
Sense  of  lines  and  surfaces,  56,  97 

Sheaf  of  lines 364 

Side 21 

Similar    systems    of    points, 

figures       ...       23,  182,  364 
Slant  height      .     .     308,  322,  323 

Small  circle 327 

Solid 1-3 

"     angle 274 

"     Geometry 245 

Sphere 326 

Spherical  cone 361 

"        excess 353 

"        polygon      ....  335 

"        sector 361 

"        segment      ....  356 

"        surface 326 

wedge 338 

Square 60,  94 

Squaring  the  circle    ....  225 

Straight  angle 5,  266 

"       line 3 

Subtend 115,  128 


PAGE 

Sum    ...       4,  5,  90,  115,  266 

Superposition 23 

Supplement      .     .     .    8,  115,  266 
Surface,  3,  21,  273,  291,  292, 

294,  317,  321,  326 
Symbols  ........     12 

Symmetry,  29,  183,  233,  275,  339 

Tables 371 

Tangent  ....     125,  143,  332 
Terms  of  a  proportion   .     .     .  161 

Tetrahedral 274 

Theorem 9 

Third  proportional     .     .     .     .195 

Touch 125,  143 

Transversal 43 

Transverse  section, 

291,  294,  317,  321 

Trapezium .59 

Trapezoid 59 

Triangle 22,  342 

Trihedral 274 

Trisect 87 

Truncated  pyramid   ....  295 

Unique 7 

Unit 112,  159,  202 

Variable 167 

Vertex,     5,  21,  23,  170,  283, 

294,  295,  321 

Vertical 8,  266 

Wedge 315,  338 

Zone 356 


UNIVmc"TY  OF  C/TJFORNTA 


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